Chapter 15 Probability

Given:

A single coin is tossed once.

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To Find:

1. The probability of getting a head.

2. The probability of getting a tail.

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Solution:

When a coin is tossed once, there are two possible outcomes: Head (H) or Tail (T).

The set of all possible outcomes is the sample space, denoted by $S$.

$S = \{H, T\}$

The total number of possible outcomes is $n(S) = 2$.

Let $E_1$ be the event of getting a head.

$E_1 = \{H\}$

The number of favorable outcomes for getting a head is $n(E_1) = 1$.

The probability of an event $E$ is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}$

The probability of getting a head is:

$P(\text{Head}) = P(E_1) = \frac{n(E_1)}{n(S)} = \frac{1}{2}$

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Let $E_2$ be the event of getting a tail.

$E_2 = \{T\}$

The number of favorable outcomes for getting a tail is $n(E_2) = 1$.

The probability of getting a tail is:

$P(\text{Tail}) = P(E_2) = \frac{n(E_2)}{n(S)} = \frac{1}{2}$

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The probability of getting a head is $\frac{1}{2}$.

The probability of getting a tail is $\frac{1}{2}$.

Given:

A bag contains:

1 red ball

1 blue ball

1 yellow ball

All balls are of the same size, so each outcome is equally likely.

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To Find:

The probability of taking out a:

(i) yellow ball

(ii) red ball

(iii) blue ball

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Solution:

The total number of balls in the bag is the sum of the number of red, blue, and yellow balls.

Total number of possible outcomes = Number of red balls + Number of blue balls + Number of yellow balls

Total number of possible outcomes = $1 + 1 + 1 = 3$

The sample space $S$ consists of the three balls: $\{R, B, Y\}$, where R represents the red ball, B the blue ball, and Y the yellow ball.

The total number of outcomes is $n(S) = 3$.

The probability of an event $E$ is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}$

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(i) Probability of taking out the yellow ball:

Let $E_1$ be the event of taking out a yellow ball.

Number of favorable outcomes for $E_1$ = Number of yellow balls = 1

$n(E_1) = 1$

$P(\text{Yellow ball}) = P(E_1) = \frac{n(E_1)}{n(S)} = \frac{1}{3}$

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(ii) Probability of taking out the red ball:

Let $E_2$ be the event of taking out a red ball.

Number of favorable outcomes for $E_2$ = Number of red balls = 1

$n(E_2) = 1$

$P(\text{Red ball}) = P(E_2) = \frac{n(E_2)}{n(S)} = \frac{1}{3}$

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(iii) Probability of taking out the blue ball:

Let $E_3$ be the event of taking out a blue ball.

Number of favorable outcomes for $E_3$ = Number of blue balls = 1

$n(E_3) = 1$

$P(\text{Blue ball}) = P(E_3) = \frac{n(E_3)}{n(S)} = \frac{1}{3}$

Given:

A die is thrown once.

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To Find:

(i) The probability of getting a number greater than 4.

(ii) The probability of getting a number less than or equal to 4.

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Solution:

When a standard die is thrown once, the possible outcomes are the numbers on its faces: 1, 2, 3, 4, 5, and 6.

The set of all possible outcomes is the sample space, $S = \{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes is $n(S) = 6$.

The probability of an event $E$ is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}$

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(i) Probability of getting a number greater than 4:

Let $E_1$ be the event of getting a number greater than 4.

The outcomes greater than 4 in the sample space are 5 and 6.

The favorable outcomes for $E_1$ are $\{5, 6\}$.

The number of favorable outcomes for $E_1$ is $n(E_1) = 2$.

The probability of getting a number greater than 4 is:

$P(\text{number} > 4) = P(E_1) = \frac{n(E_1)}{n(S)} = \frac{2}{6} = \frac{1}{3}$

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(ii) Probability of getting a number less than or equal to 4:

Let $E_2$ be the event of getting a number less than or equal to 4.

The outcomes less than or equal to 4 in the sample space are 1, 2, 3, and 4.

The favorable outcomes for $E_2$ are $\{1, 2, 3, 4\}$.

The number of favorable outcomes for $E_2$ is $n(E_2) = 4$.

The probability of getting a number less than or equal to 4 is:

$P(\text{number} \leq 4) = P(E_2) = \frac{n(E_2)}{n(S)} = \frac{4}{6} = \frac{2}{3}$

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The probability of getting a number greater than 4 is $\frac{1}{3}$.

The probability of getting a number less than or equal to 4 is $\frac{2}{3}$.

Note that the event "number greater than 4" and the event "number less than or equal to 4" are complementary events. The sum of their probabilities is $P(E_1) + P(E_2) = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1$.

Given:

One card is drawn from a well-shuffled deck of 52 cards.

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To Find:

(i) The probability that the card will be an ace.

(ii) The probability that the card will not be an ace.

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Solution:

A standard deck of 52 playing cards has the following composition:

• Suits: 4 (Hearts, Diamonds, Clubs, Spades)
• Cards in each suit: 13 (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King)

The total number of possible outcomes when drawing one card is the total number of cards in the deck.

Total number of possible outcomes = $52$.

The probability of an event $E$ is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability that the card will be an ace:

Let $E_1$ be the event that the card drawn is an ace.

In a deck of 52 cards, there are 4 aces (one in each suit: Ace of Hearts, Ace of Diamonds, Ace of Clubs, Ace of Spades).

Number of favorable outcomes for $E_1$ = Number of aces = 4.

The probability of getting an ace is:

$P(\text{an ace}) = P(E_1) = \frac{\text{Number of aces}}{\text{Total number of cards}} = \frac{4}{52}$

$P(\text{an ace}) = \frac{\cancel{4}^{1}}{\cancel{52}_{13}} = \frac{1}{13}$

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(ii) Probability that the card will not be an ace:

Let $E_2$ be the event that the card drawn is not an ace.

The event $E_2$ is the complement of the event $E_1$ (getting an ace). This means that if event $E_1$ does not happen, then event $E_2$ must happen, and vice versa.

The number of cards that are not aces = Total number of cards - Number of aces

Number of non-ace cards = $52 - 4 = 48$.

Using the probability formula:

$P(\text{not an ace}) = P(E_2) = \frac{\text{Number of non-ace cards}}{\text{Total number of cards}} = \frac{48}{52}$

$P(\text{not an ace}) = \frac{\cancel{48}^{12}}{\cancel{52}_{13}} = \frac{12}{13}$

Alternatively, using the complement rule, $P(E_2) = P(E_1') = 1 - P(E_1)$:

$P(\text{not an ace}) = 1 - P(\text{an ace})$

$P(\text{not an ace}) = 1 - \frac{1}{13}$

$P(\text{not an ace}) = \frac{13}{13} - \frac{1}{13} = \frac{13 - 1}{13} = \frac{12}{13}$

Both methods give the same result.

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The probability of getting an ace is $\frac{1}{13}$.

The probability of not getting an ace is $\frac{12}{13}$.

Given:

Probability of Sangeeta winning the match, $P(\text{Sangeeta}) = 0.62$.

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To Find:

The probability of Reshma winning the match, $P(\text{Reshma})$.

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Solution:

Let S be the event that Sangeeta wins the match.

Let R be the event that Reshma wins the match.

In a tennis match between two players, there are only two possible outcomes: either Sangeeta wins or Reshma wins. These two events are mutually exclusive and exhaustive.

The sum of the probabilities of all possible outcomes in an experiment is equal to 1.

Therefore, we can write:

$P(\text{Sangeeta}) + P(\text{Reshma}) = 1$

We are given the probability of Sangeeta winning:

$P(\text{Sangeeta}) = 0.62$

Substitute this value into the equation:

$0.62 + P(\text{Reshma}) = 1$

To find the probability of Reshma winning, subtract 0.62 from both sides of the equation:

$P(\text{Reshma}) = 1 - 0.62$

Calculating the difference:

$P(\text{Reshma}) = 0.38$

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Answer:

The probability of Reshma winning the match is $\mathbf{0.38}$.

Given:

Savita and Hamida are friends. We need to find the probabilities related to their birthdays, ignoring a leap year.

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To Find:

(i) The probability that both will have different birthdays.

(ii) The probability that both will have the same birthday.

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Solution:

Assuming a non-leap year, the total number of days in a year is 365.

The birthday of Savita can be any one of these 365 days.

The birthday of Hamida can also be any one of these 365 days.

The total number of possible outcomes for the birthdays of both Savita and Hamida is the product of the number of choices for each person's birthday.

Total number of possible outcomes = $365 \times 365 = 365^2$.

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(i) Probability of different birthdays:

For Savita and Hamida to have different birthdays, Savita's birthday can be any of the 365 days.

Then, Hamida's birthday must be one of the remaining 364 days (any day except Savita's birthday).

Number of favourable outcomes for different birthdays = $365 \times 364$.

The probability that both have different birthdays is:

$P(\text{different birthdays}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{different birthdays}) = \frac{365 \times 364}{365 \times 365}$

$P(\text{different birthdays}) = \frac{\cancel{365} \times 364}{\cancel{365} \times 365}$

$P(\text{different birthdays}) = \frac{364}{365}$

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(ii) Probability of the same birthday:

For Savita and Hamida to have the same birthday, Savita's birthday can be any of the 365 days.

Then, Hamida's birthday must be the same day as Savita's birthday. There is only 1 choice for Hamida's birthday for each choice of Savita's birthday.

Number of favourable outcomes for the same birthday = $365 \times 1 = 365$.

The probability that both have the same birthday is:

$P(\text{same birthday}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{same birthday}) = \frac{365}{365 \times 365}$

$P(\text{same birthday}) = \frac{\cancel{365}}{365 \times \cancel{365}}$

$P(\text{same birthday}) = \frac{1}{365}$

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Alternate Solution for (ii):

The events "having different birthdays" and "having the same birthday" are complementary events. This means that one event must happen if the other does not.

For any event A, the probability of the event not happening is $P(\text{not } A) = 1 - P(A)$.

Here, $P(\text{same birthday}) = 1 - P(\text{different birthdays})$.

From part (i), we found $P(\text{different birthdays}) = \frac{364}{365}$.

So, $P(\text{same birthday}) = 1 - \frac{364}{365}$

$P(\text{same birthday}) = \frac{365}{365} - \frac{364}{365}$

$P(\text{same birthday}) = \frac{365 - 364}{365}$

$P(\text{same birthday}) = \frac{1}{365}$

This confirms the result obtained earlier.

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Answer:

(i) The probability that both will have different birthdays is $\mathbf{\frac{364}{365}}$.

(ii) The probability that both will have the same birthday is $\mathbf{\frac{1}{365}}$.

Given:

Total number of students in Class X = 40.

Number of girls = 25.

Number of boys = 15.

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To Find:

(i) The probability that the selected student is a girl.

(ii) The probability that the selected student is a boy.

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Solution:

The teacher selects one student randomly from the bag containing cards with the names of all 40 students.

The total number of possible outcomes is the total number of students, which is 40.

Total number of possible outcomes = 40.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability of selecting a girl:

The number of favourable outcomes for selecting a girl is the number of girls in the class.

Number of favourable outcomes (girls) = 25.

The probability of selecting a girl is:

$P(\text{girl}) = \frac{\text{Number of girls}}{\text{Total number of students}}$

$P(\text{girl}) = \frac{25}{40}$

We can simplify this fraction by dividing the numerator and the denominator by their greatest common divisor, which is 5.

$P(\text{girl}) = \frac{\cancel{25}^{5}}{\cancel{40}_{8}}$

$P(\text{girl}) = \frac{5}{8}$

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(ii) Probability of selecting a boy:

The number of favourable outcomes for selecting a boy is the number of boys in the class.

Number of favourable outcomes (boys) = 15.

The probability of selecting a boy is:

$P(\text{boy}) = \frac{\text{Number of boys}}{\text{Total number of students}}$

$P(\text{boy}) = \frac{15}{40}$

We can simplify this fraction by dividing the numerator and the denominator by their greatest common divisor, which is 5.

$P(\text{boy}) = \frac{\cancel{15}^{3}}{\cancel{40}_{8}}$

$P(\text{boy}) = \frac{3}{8}$

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Alternate Solution for (ii):

The events "selecting a girl" and "selecting a boy" are complementary events, as one of these two events must occur when a student is selected, and they cannot occur simultaneously.

For complementary events, the sum of their probabilities is 1.

$P(\text{girl}) + P(\text{boy}) = 1$

We found $P(\text{girl}) = \frac{5}{8}$ in part (i).

So, $\frac{5}{8} + P(\text{boy}) = 1$

Subtract $\frac{5}{8}$ from both sides:

$P(\text{boy}) = 1 - \frac{5}{8}$

$P(\text{boy}) = \frac{8}{8} - \frac{5}{8}$

$P(\text{boy}) = \frac{8-5}{8}$

$P(\text{boy}) = \frac{3}{8}$

This result matches the one obtained earlier.

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Answer:

(i) The probability that the name written on the card is the name of a girl is $\mathbf{\frac{5}{8}}$.

(ii) The probability that the name written on the card is the name of a boy is $\mathbf{\frac{3}{8}}$.

Given:

Number of blue marbles = 3.

Number of white marbles = 2.

Number of red marbles = 4.

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To Find:

(i) Probability of drawing a white marble.

(ii) Probability of drawing a blue marble.

(iii) Probability of drawing a red marble.

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Solution:

First, calculate the total number of marbles in the box.

Total number of marbles = Number of blue marbles + Number of white marbles + Number of red marbles

Total number of marbles = $3 + 2 + 4 = 9$.

When a marble is drawn at random, the total number of possible outcomes is the total number of marbles, which is 9.

Total number of possible outcomes = 9.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability of drawing a white marble:

The number of favourable outcomes for drawing a white marble is the number of white marbles in the box.

Number of favourable outcomes (white) = 2.

The probability of drawing a white marble is:

$P(\text{white}) = \frac{\text{Number of white marbles}}{\text{Total number of marbles}}$

$P(\text{white}) = \frac{2}{9}$

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(ii) Probability of drawing a blue marble:

The number of favourable outcomes for drawing a blue marble is the number of blue marbles in the box.

Number of favourable outcomes (blue) = 3.

The probability of drawing a blue marble is:

$P(\text{blue}) = \frac{\text{Number of blue marbles}}{\text{Total number of marbles}}$

$P(\text{blue}) = \frac{3}{9}$

We can simplify this fraction by dividing the numerator and the denominator by their greatest common divisor, which is 3.

$P(\text{blue}) = \frac{\cancel{3}^{1}}{\cancel{9}_{3}}$

$P(\text{blue}) = \frac{1}{3}$

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(iii) Probability of drawing a red marble:

The number of favourable outcomes for drawing a red marble is the number of red marbles in the box.

Number of favourable outcomes (red) = 4.

The probability of drawing a red marble is:

$P(\text{red}) = \frac{\text{Number of red marbles}}{\text{Total number of marbles}}$

$P(\text{red}) = \frac{4}{9}$

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Check:

The events of drawing a white, blue, or red marble are mutually exclusive and exhaustive. The sum of their probabilities should be 1.

$P(\text{white}) + P(\text{blue}) + P(\text{red}) = \frac{2}{9} + \frac{1}{3} + \frac{4}{9}$

To add these fractions, find a common denominator, which is 9.

$P(\text{white}) + P(\text{blue}) + P(\text{red}) = \frac{2}{9} + \frac{1 \times 3}{3 \times 3} + \frac{4}{9}$

$P(\text{white}) + P(\text{blue}) + P(\text{red}) = \frac{2}{9} + \frac{3}{9} + \frac{4}{9}$

$P(\text{white}) + P(\text{blue}) + P(\text{red}) = \frac{2+3+4}{9} = \frac{9}{9} = 1$

The sum of probabilities is 1, which confirms our calculations are correct.

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Answer:

(i) The probability of drawing a white marble is $\mathbf{\frac{2}{9}}$.

(ii) The probability of drawing a blue marble is $\mathbf{\frac{1}{3}}$.

(iii) The probability of drawing a red marble is $\mathbf{\frac{4}{9}}$.

Given:

Harpreet tosses two different coins simultaneously (a ₹1 coin and a ₹2 coin).

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To Find:

The probability that she gets at least one head.

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Solution:

When two different coins are tossed simultaneously, the possible outcomes for the faces shown are:

Let H represent Head and T represent Tail.

Since the coins are different (₹1 and ₹2), we can list the outcomes for each coin separately.

Outcome for ₹1 coin can be H or T.

Outcome for ₹2 coin can be H or T.

The sample space S, which is the set of all possible outcomes, is:

$S = \{(\text{H on ₹1, H on ₹2}), (\text{H on ₹1, T on ₹2}), (\text{T on ₹1, H on ₹2}), (\text{T on ₹1, T on ₹2})\}$

We can represent these outcomes more concisely as:

$S = \{\text{HH, HT, TH, TT}\}$

The total number of possible outcomes is the number of elements in the sample space.

Total number of outcomes = $|S| = 4$.

We are interested in the event E, which is getting "at least one head".

This means the outcome has one or two heads.

The favourable outcomes for this event are the outcomes in the sample space that contain at least one H.

The favourable outcomes are: {HH, HT, TH}.

Number of favourable outcomes = 3.

The probability of event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{at least one head}) = \frac{3}{4}$

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Alternate Solution:

Let E be the event of getting at least one head.

The complementary event to E, denoted as E', is the event of getting "no heads".

Getting no heads means getting tails on both coins.

The outcome for E' is {TT}.

Number of outcomes for E' = 1.

The probability of event E' is:

$P(E') = P(\text{no heads}) = \frac{\text{Number of outcomes with no heads}}{\text{Total number of possible outcomes}} = \frac{1}{4}$

The probability of event E (at least one head) is $1$ minus the probability of the complementary event E'.

$P(E) = 1 - P(E')$

$P(\text{at least one head}) = 1 - \frac{1}{4}$

$P(\text{at least one head}) = \frac{4}{4} - \frac{1}{4}$

$P(\text{at least one head}) = \frac{4-1}{4}$

$P(\text{at least one head}) = \frac{3}{4}$

Both methods yield the same result.

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Answer:

The probability that Harpreet gets at least one head is $\mathbf{\frac{3}{4}}$.

Given:

The music can stop at any time within a total duration of 2 minutes after starting.

The music stops within the first half-minute after starting.

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To Find:

The probability that the music stops within the first half-minute.

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Solution:

The time at which the music stops is a random variable that can take any value within the interval from 0 minutes to 2 minutes.

The total length of the interval during which the music can stop is 2 minutes.

Total possible time interval = 2 minutes.

The event that the music stops within the first half-minute corresponds to the time interval from 0 minutes to 0.5 minutes.

Favourable time interval = 0.5 minutes.

In this type of problem involving a continuous range of time, the probability of an event is the ratio of the length of the favourable interval to the length of the total possible interval.

Probability = $\frac{\text{Length of favourable interval}}{\text{Length of total possible interval}}$

$P(\text{music stops within first 0.5 minutes}) = \frac{0.5 \text{ minutes}}{2 \text{ minutes}}$

$P(\text{music stops within first 0.5 minutes}) = \frac{0.5}{2}$

To calculate the fraction:

$\frac{0.5}{2} = \frac{1/2}{2} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

The probability can also be expressed as a decimal:

$\frac{1}{4} = 0.25$

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Answer:

The probability that the music will stop within the first half-minute after starting is $\mathbf{\frac{1}{4}}$ or $\mathbf{0.25}$.

Given:

A helicopter crashed somewhere in a rectangular region with dimensions 9 km by 4.5 km.

A lake is located within this region, which is also rectangular, with dimensions 6 km by 2 km.

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To Find:

The probability that the helicopter crashed inside the lake.

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Solution:

This is a problem of geometric probability, where the probability is determined by the ratio of areas.

The sample space is the entire rectangular region where the helicopter could have crashed.

The total area of the rectangular region is given by:

Area of rectangle = length $\times$ width

Total Area = $9 \text{ km} \times 4.5 \text{ km} = 40.5$ square km.

The favourable event is that the helicopter crashed inside the lake.

The area of the lake is given by:

Area of lake = length $\times$ width

Lake Area = $6 \text{ km} \times 2 \text{ km} = 12$ square km.

The probability that the helicopter crashed inside the lake is the ratio of the area of the lake to the total area of the rectangular region.

Probability (crashed inside lake) = $\frac{\text{Area of Lake}}{\text{Total Area}}$

Probability (crashed inside lake) = $\frac{12 \text{ km}^2}{40.5 \text{ km}^2}$

Probability (crashed inside lake) = $\frac{12}{40.5}$

To remove the decimal, multiply the numerator and denominator by 10:

Probability = $\frac{12 \times 10}{40.5 \times 10} = \frac{120}{405}$

Now, simplify the fraction. Both 120 and 405 are divisible by 5.

$\frac{120 \div 5}{405 \div 5} = \frac{24}{81}$

Both 24 and 81 are divisible by 3.

$\frac{24 \div 3}{81 \div 3} = \frac{8}{27}$

The fraction $\frac{8}{27}$ cannot be simplified further.

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Answer:

The probability that the helicopter crashed inside the lake is $\mathbf{\frac{8}{27}}$.

Given:

Total number of shirts in the carton = 100.

Number of good shirts = 88.

Number of shirts with minor defects = 8.

Number of shirts with major defects = 4.

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To Find:

(i) The probability that the drawn shirt is acceptable to Jimmy.

(ii) The probability that the drawn shirt is acceptable to Sujatha.

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Solution:

One shirt is drawn at random from the carton.

The total number of possible outcomes is the total number of shirts, which is 100.

Total number of possible outcomes = 100.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability that the shirt is acceptable to Jimmy:

Jimmy will only accept shirts which are good.

The number of favourable outcomes for this event is the number of good shirts.

Number of favourable outcomes (acceptable to Jimmy) = 88.

The probability that the shirt is acceptable to Jimmy is:

$P(\text{acceptable to Jimmy}) = \frac{\text{Number of good shirts}}{\text{Total number of shirts}}$

$P(\text{acceptable to Jimmy}) = \frac{88}{100}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4.

$P(\text{acceptable to Jimmy}) = \frac{\cancel{88}^{22}}{\cancel{100}_{25}}$

$P(\text{acceptable to Jimmy}) = \frac{22}{25}$

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(ii) Probability that the shirt is acceptable to Sujatha:

Sujatha will only reject shirts which have major defects.

This means Sujatha will accept shirts that are good or have minor defects.

Number of shirts acceptable to Sujatha = Number of good shirts + Number of shirts with minor defects

Number of shirts acceptable to Sujatha = $88 + 8 = 96$.

The number of favourable outcomes for this event is 96.

The probability that the shirt is acceptable to Sujatha is:

$P(\text{acceptable to Sujatha}) = \frac{\text{Number of shirts acceptable to Sujatha}}{\text{Total number of shirts}}$

$P(\text{acceptable to Sujatha}) = \frac{96}{100}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4.

$P(\text{acceptable to Sujatha}) = \frac{\cancel{96}^{24}}{\cancel{100}_{25}}$

$P(\text{acceptable to Sujatha}) = \frac{24}{25}$

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Answer:

(i) The probability that the shirt is acceptable to Jimmy is $\mathbf{\frac{22}{25}}$.

(ii) The probability that the shirt is acceptable to Sujatha is $\mathbf{\frac{24}{25}}$.

Given:

Two dice, one blue and one grey, are thrown simultaneously.

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To Find:

All possible outcomes and the probability that the sum of the numbers is:

(i) 8?

(ii) 13?

(iii) less than or equal to 12?

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Solution:

When two dice are thrown, the outcome can be represented as an ordered pair (number on blue die, number on grey die), where each number can be any integer from 1 to 6.

The sample space S, which is the set of all possible outcomes, is:

$S = \{$

$(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),$

$(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),$

$(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),$

$(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),$

$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),$

$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$

$\}$

The total number of possible outcomes is the number of elements in the sample space.

Total number of outcomes = $6 \times 6 = 36$.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability that the sum is 8:

Let E$_{1}$ be the event that the sum of the numbers on the two dice is 8.

The favourable outcomes are the pairs whose sum is 8:

E$_{1} = \{(2,6), (3,5), (4,4), (5,3), (6,2)\}$.

The number of favourable outcomes for E$_{1}$ is 5.

The probability of event E$_{1}$ is:

$P(E_{1}) = P(\text{sum is 8}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{sum is 8}) = \frac{5}{36}$

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(ii) Probability that the sum is 13:

Let E$_{2}$ be the event that the sum of the numbers on the two dice is 13.

The maximum possible sum when rolling two dice is $6 + 6 = 12$.

Therefore, there are no outcomes where the sum is 13.

The set of favourable outcomes for E$_{2}$ is empty, E$_{2} = \{\}$.

The number of favourable outcomes for E$_{2}$ is 0.

The probability of event E$_{2}$ is:

$P(E_{2}) = P(\text{sum is 13}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{sum is 13}) = \frac{0}{36} = 0$

This is an impossible event.

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(iii) Probability that the sum is less than or equal to 12:

Let E$_{3}$ be the event that the sum of the numbers on the two dice is less than or equal to 12.

The minimum possible sum is $1 + 1 = 2$.

The maximum possible sum is $6 + 6 = 12$.

All possible outcomes from the sample space S have a sum between 2 and 12, inclusive.

Therefore, all 36 outcomes in the sample space are favourable outcomes for E$_{3}$.

The number of favourable outcomes for E$_{3}$ is 36.

The probability of event E$_{3}$ is:

$P(E_{3}) = P(\text{sum} \leq 12) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{sum} \leq 12) = \frac{36}{36} = 1$

This is a certain event.

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Answer:

The possible outcomes when throwing two dice are all the pairs $(b, g)$ where $b, g \in \{1, 2, 3, 4, 5, 6\}$. There are 36 possible outcomes.

(i) The probability that the sum of the two numbers appearing on the top of the dice is 8 is $\mathbf{\frac{5}{36}}$.

(ii) The probability that the sum of the two numbers appearing on the top of the dice is 13 is $\mathbf{0}$.

(iii) The probability that the sum of the two numbers appearing on the top of the dice is less than or equal to 12 is $\mathbf{1}$.

(i) Probability of an event E + Probability of the event ‘not E’ = $\mathbf{1}$.

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(ii) The probability of an event that cannot happen is $\mathbf{0}$. Such an event is called $\mathbf{impossible \ event}$.

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(iii) The probability of an event that is certain to happen is $\mathbf{1}$. Such an event is called $\mathbf{sure \ event}$ or $\mathbf{certain \ event}$.

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(iv) The sum of the probabilities of all the elementary events of an experiment is $\mathbf{1}$.

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(v) The probability of an event is greater than or equal to $\mathbf{0}$ and less than or equal to $\mathbf{1}$.

(i) Not equally likely outcomes.

The probability of a car starting or not starting depends on various factors such as the car's condition, fuel level, battery status, etc. These factors can make one outcome (starting) more or less likely than the other (not starting).

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(ii) Not equally likely outcomes.

The probability of a player shooting or missing a basketball shot depends on the player's skill level, distance from the basket, presence of defenders, etc. A skilled player is more likely to make the shot than a beginner, so the outcomes are not equally likely in general.

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(iii) Equally likely outcomes.

Assuming the question is a standard true-false question and there is no prior knowledge influencing the answer, there are only two possible outcomes: the answer is either right or wrong. When a trial is made (implying a choice or guess without bias), each outcome has an equal chance of occurring.

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(iv) Equally likely outcomes.

While there are minor biological variations that cause a slight difference in the birth rate of boys and girls globally, for the purpose of a standard probability problem at this level, the outcomes of a baby being born a boy or a girl are considered to be equally likely. Each outcome has a probability of approximately $\frac{1}{2}$.

Tossing a coin is considered a fair way of deciding which team should get the ball at the beginning of a football game because the outcome of a coin toss is **equally likely** for both possibilities.

When a standard, unbiased coin is tossed, there are only two possible outcomes: getting a Head (H) or getting a Tail (T).

The probability of getting a Head is $\frac{1}{2}$.

The probability of getting a Tail is $\frac{1}{2}$.

Since $P(\text{Head}) = P(\text{Tail}) = \frac{1}{2}$, both outcomes are equally likely. This means that each team has an equal chance of winning the toss (assuming one team chooses Heads and the other chooses Tails).

Furthermore, the outcome of a coin toss is unpredictable and random, as there is no way to influence or predict the result of a single toss.

Because each team has an equal probability of winning the toss, it provides a neutral and unbiased method for making the decision, ensuring fairness at the start of the game.

Solution:

The probability of any event E, denoted by $P(E)$, must satisfy the condition:

$0 \leq P(E) \leq 1$

Let's examine each option:

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(A) $\frac{2}{3}$

This is a fraction. $\frac{2}{3} \approx 0.667$.

Since $0 \leq 0.667 \leq 1$, this value can be a probability.

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(B) –1.5

This is a negative number.

The probability of an event cannot be negative.

Since $-1.5 < 0$, this value cannot be a probability.

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(C) 15%

This is a percentage. We can convert it to a decimal:

$15\% = \frac{15}{100} = 0.15$

Since $0 \leq 0.15 \leq 1$, this value can be a probability.

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(D) 0.7

This is a decimal number.

Since $0 \leq 0.7 \leq 1$, this value can be a probability.

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Based on the property that the probability of an event must be between 0 and 1 (inclusive), the value –1.5 is the only option that falls outside this range.

---

Answer:

The option that cannot be the probability of an event is $\mathbf{(B) \ –1.5}$.

Given:

The probability of event E is $P(E) = 0.05$.

---

To Find:

The probability of the event 'not E', often denoted as $P(\text{not } E)$ or $P(\bar{E})$.

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Solution:

For any event E, the event 'not E' is the complementary event. This means that event E and event 'not E' are mutually exclusive and exhaustive.

The sum of the probability of an event and the probability of its complementary event is always equal to 1.

The relationship is given by the formula:

$P(E) + P(\text{not } E) = 1$

We are given that $P(E) = 0.05$. Substitute this value into the formula:

$0.05 + P(\text{not } E) = 1$

To find $P(\text{not } E)$, subtract 0.05 from both sides of the equation:

$P(\text{not } E) = 1 - 0.05$

Calculating the subtraction:

$P(\text{not } E) = 0.95$

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Answer:

The probability of ‘not E’ is $\mathbf{0.95}$.

Given:

A bag contains only lemon flavoured candies.

Malini takes out one candy at random.

---

To Find:

(i) The probability that she takes out an orange flavoured candy.

(ii) The probability that she takes out a lemon flavoured candy.

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Solution:

Let $N$ be the total number of candies in the bag. Since the bag contains only lemon flavoured candies, all $N$ candies are lemon flavoured.

Total number of possible outcomes when drawing one candy = $N$.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability of taking out an orange flavoured candy:

The event is drawing an orange flavoured candy.

Since the bag contains only lemon flavoured candies, there are no orange flavoured candies in the bag.

Number of favourable outcomes (orange) = 0.

The probability of drawing an orange flavoured candy is:

$P(\text{orange}) = \frac{\text{Number of orange candies}}{\text{Total number of candies}}$

$P(\text{orange}) = \frac{0}{N}$

$P(\text{orange}) = 0$

This is an impossible event.

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(ii) Probability of taking out a lemon flavoured candy:

The event is drawing a lemon flavoured candy.

Since the bag contains only lemon flavoured candies, every candy in the bag is a lemon flavoured candy.

Number of favourable outcomes (lemon) = Total number of candies = $N$.

The probability of drawing a lemon flavoured candy is:

$P(\text{lemon}) = \frac{\text{Number of lemon candies}}{\text{Total number of candies}}$

$P(\text{lemon}) = \frac{N}{N}$

$P(\text{lemon}) = 1$

This is a certain event.

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Answer:

(i) The probability that she takes out an orange flavoured candy is $\mathbf{0}$.

(ii) The probability that she takes out a lemon flavoured candy is $\mathbf{1}$.

Given:

The probability that 2 students do not have the same birthday, $P(\text{not having same birthday}) = 0.992$.

---

To Find:

The probability that the 2 students have the same birthday.

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Solution:

Let E be the event that 2 students have the same birthday.

Let $\bar{E}$ be the event that 2 students do not have the same birthday.

The events E and $\bar{E}$ are complementary events. This means that the sum of their probabilities is 1.

$P(E) + P(\bar{E}) = 1$

We are given $P(\bar{E}) = 0.992$.

Substitute the given value into the equation:

$P(E) + 0.992 = 1$

To find $P(E)$, subtract 0.992 from both sides of the equation:

$P(E) = 1 - 0.992$

Calculating the subtraction:

$P(E) = 0.008$

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Answer:

The probability that the 2 students have the same birthday is $\mathbf{0.008}$.

Given:

Number of red balls = 3.

Number of black balls = 5.

A ball is drawn at random from the bag.

---

To Find:

(i) The probability that the ball drawn is red.

(ii) The probability that the ball drawn is not red.

---

Solution:

First, calculate the total number of balls in the bag.

Total number of balls = Number of red balls + Number of black balls

Total number of balls = $3 + 5 = 8$.

When a ball is drawn at random, the total number of possible outcomes is the total number of balls, which is 8.

Total number of possible outcomes = 8.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability that the ball drawn is red:

The event is drawing a red ball.

The number of favourable outcomes for this event is the number of red balls.

Number of favourable outcomes (red) = 3.

The probability of drawing a red ball is:

$P(\text{red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}}$

$P(\text{red}) = \frac{3}{8}$

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(ii) Probability that the ball drawn is not red:

The event 'not red' means drawing a ball that is not red.

In this bag, the balls that are not red are the black balls.

Number of favourable outcomes (not red) = Number of black balls = 5.

The probability of drawing a ball that is not red is:

$P(\text{not red}) = \frac{\text{Number of black balls}}{\text{Total number of balls}}$

$P(\text{not red}) = \frac{5}{8}$

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Alternate Solution for (ii):

The event "drawing a red ball" and the event "drawing a ball that is not red" are complementary events. This means that one event must happen if the other does not.

For any event E, $P(\text{not } E) = 1 - P(E)$.

Here, $P(\text{not red}) = 1 - P(\text{red})$.

From part (i), we found $P(\text{red}) = \frac{3}{8}$.

So, $P(\text{not red}) = 1 - \frac{3}{8}$

$P(\text{not red}) = \frac{8}{8} - \frac{3}{8}$

$P(\text{not red}) = \frac{8-3}{8}$

$P(\text{not red}) = \frac{5}{8}$

This confirms the result obtained earlier.

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Answer:

(i) The probability that the ball drawn is red is $\mathbf{\frac{3}{8}}$.

(ii) The probability that the ball drawn is not red is $\mathbf{\frac{5}{8}}$.

Given:

Number of red marbles = 5.

Number of white marbles = 8.

Number of green marbles = 4.

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To Find:

(i) Probability of drawing a red marble.

(ii) Probability of drawing a white marble.

(iii) Probability of drawing a marble that is not green.

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Solution:

First, calculate the total number of marbles in the box.

Total number of marbles = Number of red marbles + Number of white marbles + Number of green marbles

Total number of marbles = $5 + 8 + 4 = 17$.

When a marble is taken out at random, the total number of possible outcomes is the total number of marbles, which is 17.

Total number of possible outcomes = 17.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability of drawing a red marble:

The event is drawing a red marble.

The number of favourable outcomes for this event is the number of red marbles.

Number of favourable outcomes (red) = 5.

The probability of drawing a red marble is:

$P(\text{red}) = \frac{\text{Number of red marbles}}{\text{Total number of marbles}}$

$P(\text{red}) = \frac{5}{17}$

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(ii) Probability of drawing a white marble:

The event is drawing a white marble.

The number of favourable outcomes for this event is the number of white marbles.

Number of favourable outcomes (white) = 8.

The probability of drawing a white marble is:

$P(\text{white}) = \frac{\text{Number of white marbles}}{\text{Total number of marbles}}$

$P(\text{white}) = \frac{8}{17}$

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(iii) Probability of drawing a marble that is not green:

The event 'not green' means drawing a marble that is either red or white.

The number of marbles that are not green = Number of red marbles + Number of white marbles

Number of favourable outcomes (not green) = $5 + 8 = 13$.

The probability of drawing a marble that is not green is:

$P(\text{not green}) = \frac{\text{Number of marbles that are not green}}{\text{Total number of marbles}}$

$P(\text{not green}) = \frac{13}{17}$

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Alternate Solution for (iii):

The event "drawing a green marble" and the event "drawing a marble that is not green" are complementary events.

First, calculate the probability of drawing a green marble:

$P(\text{green}) = \frac{\text{Number of green marbles}}{\text{Total number of marbles}} = \frac{4}{17}$

Now, use the complementary event formula: $P(\text{not green}) = 1 - P(\text{green})$.

$P(\text{not green}) = 1 - \frac{4}{17}$

$P(\text{not green}) = \frac{17}{17} - \frac{4}{17}$

$P(\text{not green}) = \frac{17-4}{17}$

$P(\text{not green}) = \frac{13}{17}$

This confirms the result obtained earlier.

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Answer:

(i) The probability that the marble taken out will be red is $\mathbf{\frac{5}{17}}$.

(ii) The probability that the marble taken out will be white is $\mathbf{\frac{8}{17}}$.

(iii) The probability that the marble taken out will be not green is $\mathbf{\frac{13}{17}}$.

Given:

Number of 50p coins = 100.

Number of $\textsf{₹}$1 coins = 50.

Number of $\textsf{₹}$2 coins = 20.

Number of $\textsf{₹}$5 coins = 10.

---

To Find:

(i) Probability that the coin is a 50p coin.

(ii) Probability that the coin is not a $\textsf{₹}$5 coin.

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Solution:

First, calculate the total number of coins in the piggy bank.

Total number of coins = (Number of 50p coins) + (Number of $\textsf{₹}$1 coins) + (Number of $\textsf{₹}$2 coins) + (Number of $\textsf{₹}$5 coins)

Total number of coins = $100 + 50 + 20 + 10 = 180$.

When one coin falls out, the total number of possible outcomes is 180.

Total number of possible outcomes = 180.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability that the coin will be a 50p coin:

The event is getting a 50p coin.

The number of favourable outcomes for this event is the number of 50p coins.

Number of favourable outcomes (50p) = 100.

The probability of getting a 50p coin is:

$P(\text{50p}) = \frac{\text{Number of 50p coins}}{\text{Total number of coins}}$

$P(\text{50p}) = \frac{100}{180}$

Simplify the fraction by dividing numerator and denominator by their greatest common divisor, which is 20.

$P(\text{50p}) = \frac{\cancel{100}^{5}}{\cancel{180}_{9}}$

$P(\text{50p}) = \frac{5}{9}$

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(ii) Probability that the coin will not be a $\textsf{₹}$5 coin:

The event is getting a coin that is not a $\textsf{₹}$5 coin.

The coins that are not $\textsf{₹}$5 are the 50p, $\textsf{₹}$1, and $\textsf{₹}$2 coins.

Number of coins that are not $\textsf{₹}$5 = (Number of 50p coins) + (Number of $\textsf{₹}$1 coins) + (Number of $\textsf{₹}$2 coins)

Number of favourable outcomes (not $\textsf{₹}$5) = $100 + 50 + 20 = 170$.

The probability of getting a coin that is not a $\textsf{₹}$5 coin is:

$P(\text{not } \textsf{₹}5) = \frac{\text{Number of coins not } \textsf{₹}5}{\text{Total number of coins}}$

$P(\text{not } \textsf{₹}5) = \frac{170}{180}$

Simplify the fraction by dividing numerator and denominator by their greatest common divisor, which is 10.

$P(\text{not } \textsf{₹}5) = \frac{\cancel{170}^{17}}{\cancel{180}_{18}}$

$P(\text{not } \textsf{₹}5) = \frac{17}{18}$

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Alternate Solution for (ii):

The event "getting a $\textsf{₹}$5 coin" and the event "getting a coin that is not a $\textsf{₹}$5 coin" are complementary events.

First, calculate the probability of getting a $\textsf{₹}$5 coin:

$P(\textsf{₹}5) = \frac{\text{Number of } \textsf{₹}5 \text{ coins}}{\text{Total number of coins}} = \frac{10}{180} = \frac{1}{18}$

Now, use the complementary event formula: $P(\text{not } \textsf{₹}5) = 1 - P(\textsf{₹}5)$.

$P(\text{not } \textsf{₹}5) = 1 - \frac{1}{18}$

$P(\text{not } \textsf{₹}5) = \frac{18}{18} - \frac{1}{18}$

$P(\text{not } \textsf{₹}5) = \frac{18-1}{18}$

$P(\text{not } \textsf{₹}5) = \frac{17}{18}$

This confirms the result obtained earlier.

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Answer:

(i) The probability that the coin will be a 50p coin is $\mathbf{\frac{5}{9}}$.

(ii) The probability that the coin will not be a $\textsf{₹}$5 coin is $\mathbf{\frac{17}{18}}$.

Given:

Number of male fish = 5.

Number of female fish = 8.

---

To Find:

The probability that the fish taken out is a male fish.

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Solution:

First, calculate the total number of fish in the tank.

Total number of fish = Number of male fish + Number of female fish

Total number of fish = $5 + 8 = 13$.

When one fish is taken out at random, the total number of possible outcomes is the total number of fish, which is 13.

Total number of possible outcomes = 13.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

The event is drawing a male fish.

The number of favourable outcomes for this event is the number of male fish.

Number of favourable outcomes (male) = 5.

The probability of drawing a male fish is:

$P(\text{male fish}) = \frac{\text{Number of male fish}}{\text{Total number of fish}}$

$P(\text{male fish}) = \frac{5}{13}$

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Answer:

The probability that the fish taken out is a male fish is $\mathbf{\frac{5}{13}}$.

Given:

An arrow is spun and comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8.

The outcomes are equally likely.

---

To Find:

The probability that the arrow will point at:

(i) 8

(ii) an odd number

(iii) a number greater than 2

(iv) a number less than 9

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Solution:

The set of all possible outcomes when the arrow stops is the sample space S:

$S = \{1, 2, 3, 4, 5, 6, 7, 8\}$

The total number of possible outcomes is the number of elements in the sample space.

Total number of outcomes = $|S| = 8$.

Since the outcomes are equally likely, the probability of any specific outcome is $\frac{1}{8}$.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability that the arrow will point at 8:

Let E$_{1}$ be the event that the arrow points at 8.

The favourable outcome for this event is {8}.

Number of favourable outcomes for E$_{1}$ = 1.

The probability of event E$_{1}$ is:

$P(E_{1}) = P(\text{pointing at 8}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{pointing at 8}) = \frac{1}{8}$

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(ii) Probability that the arrow will point at an odd number:

Let E$_{2}$ be the event that the arrow points at an odd number.

The odd numbers in the sample space S are {1, 3, 5, 7}.

The favourable outcomes for E$_{2}$ are {1, 3, 5, 7}.

Number of favourable outcomes for E$_{2}$ = 4.

The probability of event E$_{2}$ is:

$P(E_{2}) = P(\text{pointing at an odd number}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{pointing at an odd number}) = \frac{4}{8}$

Simplify the fraction by dividing numerator and denominator by 4.

$P(\text{pointing at an odd number}) = \frac{\cancel{4}^{1}}{\cancel{8}_{2}}$

$P(\text{pointing at an odd number}) = \frac{1}{2}$

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(iii) Probability that the arrow will point at a number greater than 2:

Let E$_{3}$ be the event that the arrow points at a number greater than 2.

The numbers in the sample space S that are greater than 2 are {3, 4, 5, 6, 7, 8}.

The favourable outcomes for E$_{3}$ are {3, 4, 5, 6, 7, 8}.

Number of favourable outcomes for E$_{3}$ = 6.

The probability of event E$_{3}$ is:

$P(E_{3}) = P(\text{pointing at a number greater than 2}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{pointing at a number greater than 2}) = \frac{6}{8}$

Simplify the fraction by dividing numerator and denominator by 2.

$P(\text{pointing at a number greater than 2}) = \frac{\cancel{6}^{3}}{\cancel{8}_{4}}$

$P(\text{pointing at a number greater than 2}) = \frac{3}{4}$

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(iv) Probability that the arrow will point at a number less than 9:

Let E$_{4}$ be the event that the arrow points at a number less than 9.

The numbers in the sample space S that are less than 9 are {1, 2, 3, 4, 5, 6, 7, 8}.

All outcomes in the sample space are less than 9.

The favourable outcomes for E$_{4}$ are {1, 2, 3, 4, 5, 6, 7, 8}.

Number of favourable outcomes for E$_{4}$ = 8.

The probability of event E$_{4}$ is:

$P(E_{4}) = P(\text{pointing at a number less than 9}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{pointing at a number less than 9}) = \frac{8}{8}$

$P(\text{pointing at a number less than 9}) = 1$

This is a certain event.

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Answer:

(i) The probability that it will point at 8 is $\mathbf{\frac{1}{8}}$.

(ii) The probability that it will point at an odd number is $\mathbf{\frac{1}{2}}$.

(iii) The probability that it will point at a number greater than 2 is $\mathbf{\frac{3}{4}}$.

(iv) The probability that it will point at a number less than 9 is $\mathbf{1}$.

Given:

A die is thrown once.

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To Find:

The probability of getting:

(i) a prime number.

(ii) a number lying between 2 and 6.

(iii) an odd number.

---

Solution:

When a standard six-sided die is thrown once, the set of all possible outcomes is the sample space S:

$S = \{1, 2, 3, 4, 5, 6\}$

The total number of possible outcomes is the number of elements in the sample space.

Total number of outcomes = $|S| = 6$.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability of getting a prime number:

Let E$_{1}$ be the event of getting a prime number.

Prime numbers in the sample space S are numbers greater than 1 that have no positive divisors other than 1 and themselves.

The prime numbers in S are {2, 3, 5}.

The favourable outcomes for E$_{1}$ are {2, 3, 5}.

Number of favourable outcomes for E$_{1}$ = 3.

The probability of event E$_{1}$ is:

$P(E_{1}) = P(\text{prime number}) = \frac{\text{Number of prime numbers}}{\text{Total number of outcomes}}$

$P(\text{prime number}) = \frac{3}{6}$

Simplify the fraction by dividing numerator and denominator by 3.

$P(\text{prime number}) = \frac{\cancel{3}^{1}}{\cancel{6}_{2}}$

$P(\text{prime number}) = \frac{1}{2}$

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(ii) Probability of getting a number lying between 2 and 6:

Let E$_{2}$ be the event of getting a number lying between 2 and 6.

The numbers in the sample space S that are strictly between 2 and 6 are {3, 4, 5}.

The favourable outcomes for E$_{2}$ are {3, 4, 5}.

Number of favourable outcomes for E$_{2}$ = 3.

The probability of event E$_{2}$ is:

$P(E_{2}) = P(\text{number between 2 and 6}) = \frac{\text{Number of outcomes between 2 and 6}}{\text{Total number of outcomes}}$

$P(\text{number between 2 and 6}) = \frac{3}{6}$

Simplify the fraction by dividing numerator and denominator by 3.

$P(\text{number between 2 and 6}) = \frac{\cancel{3}^{1}}{\cancel{6}_{2}}$

$P(\text{number between 2 and 6}) = \frac{1}{2}$

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(iii) Probability of getting an odd number:

Let E$_{3}$ be the event of getting an odd number.

The odd numbers in the sample space S are {1, 3, 5}.

The favourable outcomes for E$_{3}$ are {1, 3, 5}.

Number of favourable outcomes for E$_{3}$ = 3.

The probability of event E$_{3}$ is:

$P(E_{3}) = P(\text{odd number}) = \frac{\text{Number of odd numbers}}{\text{Total number of outcomes}}$

$P(\text{odd number}) = \frac{3}{6}$

Simplify the fraction by dividing numerator and denominator by 3.

$P(\text{odd number}) = \frac{\cancel{3}^{1}}{\cancel{6}_{2}}$

$P(\text{odd number}) = \frac{1}{2}$

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Answer:

(i) The probability of getting a prime number is $\mathbf{\frac{1}{2}}$.

(ii) The probability of getting a number lying between 2 and 6 is $\mathbf{\frac{1}{2}}$.

(iii) The probability of getting an odd number is $\mathbf{\frac{1}{2}}$.

Given:

A well-shuffled deck of 52 cards.

One card is drawn at random.

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To Find:

The probability of getting:

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

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Solution:

The total number of possible outcomes when drawing one card from a deck of 52 cards is 52.

Total number of possible outcomes = 52.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability of getting a king of red colour:

The red colours suits are Hearts ($\heartsuit$) and Diamonds ($\diamondsuit$).

The kings of red colour are the King of Hearts and the King of Diamonds.

Number of favourable outcomes (king of red colour) = 2.

The probability is:

$P(\text{king of red colour}) = \frac{\text{Number of kings of red colour}}{\text{Total number of cards}}$

$P(\text{king of red colour}) = \frac{2}{52}$

Simplify the fraction:

$P(\text{king of red colour}) = \frac{\cancel{2}^{1}}{\cancel{52}_{26}}$

$P(\text{king of red colour}) = \frac{1}{26}$

---

(ii) Probability of getting a face card:

The face cards in a deck are Jack, Queen, and King.

Each of the four suits (Hearts, Diamonds, Clubs, Spades) has 3 face cards.

Total number of face cards = $3 \times 4 = 12$.

Number of favourable outcomes (face card) = 12.

The probability is:

$P(\text{face card}) = \frac{\text{Number of face cards}}{\text{Total number of cards}}$

$P(\text{face card}) = \frac{12}{52}$

Simplify the fraction by dividing numerator and denominator by 4:

$P(\text{face card}) = \frac{\cancel{12}^{3}}{\cancel{52}_{13}}$

$P(\text{face card}) = \frac{3}{13}$

---

(iii) Probability of getting a red face card:

The red suits are Hearts and Diamonds.

The face cards in Hearts are Jack, Queen, King (3 cards).

The face cards in Diamonds are Jack, Queen, King (3 cards).

Total number of red face cards = $3 + 3 = 6$.

Number of favourable outcomes (red face card) = 6.

The probability is:

$P(\text{red face card}) = \frac{\text{Number of red face cards}}{\text{Total number of cards}}$

$P(\text{red face card}) = \frac{6}{52}$

Simplify the fraction by dividing numerator and denominator by 2:

$P(\text{red face card}) = \frac{\cancel{6}^{3}}{\cancel{52}_{26}}$

$P(\text{red face card}) = \frac{3}{26}$

---

(iv) Probability of getting the jack of hearts:

There is only one Jack of Hearts in a deck of 52 cards.

Number of favourable outcomes (jack of hearts) = 1.

The probability is:

$P(\text{jack of hearts}) = \frac{\text{Number of jack of hearts}}{\text{Total number of cards}}$

$P(\text{jack of hearts}) = \frac{1}{52}$

---

(v) Probability of getting a spade:

There are 13 cards of the Spade suit in a deck (Ace, 2, 3, ..., 10, Jack, Queen, King of Spades).

Number of favourable outcomes (a spade) = 13.

The probability is:

$P(\text{a spade}) = \frac{\text{Number of spades}}{\text{Total number of cards}}$

$P(\text{a spade}) = \frac{13}{52}$

Simplify the fraction by dividing numerator and denominator by 13:

$P(\text{a spade}) = \frac{\cancel{13}^{1}}{\cancel{52}_{4}}$

$P(\text{a spade}) = \frac{1}{4}$

---

(vi) Probability of getting the queen of diamonds:

There is only one Queen of Diamonds in a deck of 52 cards.

Number of favourable outcomes (queen of diamonds) = 1.

The probability is:

$P(\text{queen of diamonds}) = \frac{\text{Number of queen of diamonds}}{\text{Total number of cards}}$

$P(\text{queen of diamonds}) = \frac{1}{52}$

---

Answer:

(i) The probability of getting a king of red colour is $\mathbf{\frac{1}{26}}$.

(ii) The probability of getting a face card is $\mathbf{\frac{3}{13}}$.

(iii) The probability of getting a red face card is $\mathbf{\frac{3}{26}}$.

(iv) The probability of getting the jack of hearts is $\mathbf{\frac{1}{52}}$.

(v) The probability of getting a spade is $\mathbf{\frac{1}{4}}$.

(vi) The probability of getting the queen of diamonds is $\mathbf{\frac{1}{52}}$.

Given:

The initial set of cards is the ten, jack, queen, king, and ace of diamonds.

These 5 cards are well-shuffled, and one card is picked at random.

---

To Find:

(i) The probability that the first card is the queen.

(ii) If the queen is drawn and put aside, the probability that the second card picked up is (a) an ace? (b) a queen?

---

Solution:

The initial set of cards is $\{$10 of Diamonds, Jack of Diamonds, Queen of Diamonds, King of Diamonds, Ace of Diamonds$\}$.

The total number of cards initially is 5.

Total number of possible outcomes = 5.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

---

(i) Probability that the card is the queen:

Let E$_{1}$ be the event that the first card drawn is the queen.

There is only one Queen of Diamonds in the initial set of 5 cards.

Number of favourable outcomes for E$_{1}$ (queen) = 1.

The probability of drawing the queen is:

$P(E_{1}) = P(\text{queen}) = \frac{\text{Number of queens}}{\text{Total number of cards}}$

$P(\text{queen}) = \frac{1}{5}$

---

(ii) If the queen is drawn and put aside:

After drawing the queen and putting it aside, there are now only 4 cards left.

The remaining cards are $\{$10 of Diamonds, Jack of Diamonds, King of Diamonds, Ace of Diamonds$\}$.

The new total number of possible outcomes for the second draw is 4.

---

(ii) (a) Probability that the second card picked up is an ace:

Let E$_{2a}$ be the event that the second card drawn is an ace.

From the remaining 4 cards, there is one Ace of Diamonds.

Number of favourable outcomes for E$_{2a}$ (ace) = 1.

The probability of drawing an ace as the second card (given the queen was drawn first and put aside) is:

$P(E_{2a}) = P(\text{ace second}) = \frac{\text{Number of aces remaining}}{\text{Total number of cards remaining}}$

$P(\text{ace second}) = \frac{1}{4}$

---

(ii) (b) Probability that the second card picked up is a queen:

Let E$_{2b}$ be the event that the second card drawn is a queen.

Since the queen was drawn first and put aside, there are no queens left among the remaining 4 cards.

Number of favourable outcomes for E$_{2b}$ (queen) = 0.

The probability of drawing a queen as the second card (given the queen was drawn first and put aside) is:

$P(E_{2b}) = P(\text{queen second}) = \frac{\text{Number of queens remaining}}{\text{Total number of cards remaining}}$

$P(\text{queen second}) = \frac{0}{4}$

$P(\text{queen second}) = 0$

---

Answer:

(i) The probability that the card is the queen is $\mathbf{\frac{1}{5}}$.

(ii) If the queen is drawn and put aside:

(a) The probability that the second card picked up is an ace is $\mathbf{\frac{1}{4}}$.

(b) The probability that the second card picked up is a queen is $\mathbf{0}$.

Given:

Number of defective pens = 12.

Number of good pens = 132.

One pen is taken out at random from the lot.

---

To Find:

The probability that the pen taken out is a good one.

---

Solution:

First, calculate the total number of pens in the lot.

Total number of pens = Number of defective pens + Number of good pens

Total number of pens = $12 + 132 = 144$.

When one pen is taken out at random, the total number of possible outcomes is the total number of pens, which is 144.

Total number of possible outcomes = 144.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

The event is drawing a good pen.

The number of favourable outcomes for this event is the number of good pens.

Number of favourable outcomes (good pen) = 132.

The probability of drawing a good pen is:

$P(\text{good pen}) = \frac{\text{Number of good pens}}{\text{Total number of pens}}$

$P(\text{good pen}) = \frac{132}{144}$

Simplify the fraction. Both 132 and 144 are divisible by 12.

$\frac{132 \div 12}{144 \div 12} = \frac{11}{12}$

$P(\text{good pen}) = \frac{11}{12}$

---

Answer:

The probability that the pen taken out is a good one is $\mathbf{\frac{11}{12}}$.

Given:

Total number of bulbs in the lot = 20.

Number of defective bulbs = 4.

---

To Find:

(i) The probability that a randomly drawn bulb is defective.

(ii) The probability that a second randomly drawn bulb (from the rest, after a non-defective bulb was drawn first and not replaced) is not defective.

---

Solution:

Number of good bulbs = Total bulbs - Number of defective bulbs

Number of good bulbs = $20 - 4 = 16$.

Total number of possible outcomes when drawing one bulb = 20.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

---

(i) Probability that the first bulb drawn is defective:

The event is drawing a defective bulb.

Number of favourable outcomes (defective) = 4.

The probability of drawing a defective bulb is:

$P(\text{defective}) = \frac{\text{Number of defective bulbs}}{\text{Total number of bulbs}}$

$P(\text{defective}) = \frac{4}{20}$

Simplify the fraction by dividing numerator and denominator by 4.

$P(\text{defective}) = \frac{\cancel{4}^{1}}{\cancel{20}_{5}}$

$P(\text{defective}) = \frac{1}{5}$

---

(ii) Probability that the second bulb is not defective (given the first was not defective and not replaced):

In the first draw, a bulb that is *not* defective (i.e., a good bulb) was drawn, and it was not replaced.

Initial number of good bulbs = 16.

After drawing one good bulb, the number of good bulbs remaining = $16 - 1 = 15$.

The number of defective bulbs remains the same = 4.

The total number of bulbs remaining in the lot = (Total initial bulbs) - 1

Total number of bulbs remaining = $20 - 1 = 19$.

Now, a second bulb is drawn at random from these 19 bulbs.

The total number of possible outcomes for the second draw is 19.

The event is that this second bulb is not defective.

The number of favourable outcomes for this event is the number of good bulbs remaining.

Number of favourable outcomes (not defective in second draw) = 15.

The probability of drawing a non-defective bulb in the second draw (given the condition) is:

$P(\text{not defective second} | \text{not defective first}) = \frac{\text{Number of good bulbs remaining}}{\text{Total number of bulbs remaining}}$

$P(\text{not defective second} | \text{not defective first}) = \frac{15}{19}$

---

Answer:

(i) The probability that the first bulb drawn is defective is $\mathbf{\frac{1}{5}}$.

(ii) The probability that the second bulb is not defective (given the first was not defective and not replaced) is $\mathbf{\frac{15}{19}}$.

Given:

A box contains 90 discs numbered from 1 to 90.

One disc is drawn at random.

---

To Find:

The probability that the disc bears:

(i) a two-digit number.

(ii) a perfect square number.

(iii) a number divisible by 5.

---

Solution:

The numbers on the discs are $\{1, 2, 3, \dots, 90\}$.

The total number of possible outcomes when drawing one disc is 90.

Total number of possible outcomes = 90.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

---

(i) Probability of drawing a two-digit number:

The numbers from 1 to 9 are single-digit numbers.

Number of single-digit numbers = 9.

The numbers from 10 to 90 are two-digit numbers.

Number of two-digit numbers = Total numbers - Number of single-digit numbers

Number of two-digit numbers = $90 - 9 = 81$.

Number of favourable outcomes (two-digit number) = 81.

The probability of drawing a two-digit number is:

$P(\text{two-digit number}) = \frac{\text{Number of two-digit numbers}}{\text{Total number of discs}}$

$P(\text{two-digit number}) = \frac{81}{90}$

Simplify the fraction by dividing numerator and denominator by 9.

$P(\text{two-digit number}) = \frac{\cancel{81}^{9}}{\cancel{90}_{10}}$

$P(\text{two-digit number}) = \frac{9}{10}$

---

(ii) Probability of drawing a perfect square number:

A perfect square number is an integer that is the square of an integer.

The perfect square numbers between 1 and 90 are:

$1^2=1$

$2^2=4$

$3^2=9$

$4^2=16$

$5^2=25$

$6^2=36$

$7^2=49$

$8^2=64$

$9^2=81$

The next perfect square is $10^2=100$, which is greater than 90.

The perfect square numbers are $\{1, 4, 9, 16, 25, 36, 49, 64, 81\}$.

Number of favourable outcomes (perfect square) = 9.

The probability of drawing a perfect square number is:

$P(\text{perfect square}) = \frac{\text{Number of perfect squares}}{\text{Total number of discs}}$

$P(\text{perfect square}) = \frac{9}{90}$

Simplify the fraction by dividing numerator and denominator by 9.

$P(\text{perfect square}) = \frac{\cancel{9}^{1}}{\cancel{90}_{10}}$

$P(\text{perfect square}) = \frac{1}{10}$

---

(iii) Probability of drawing a number divisible by 5:

The numbers between 1 and 90 that are divisible by 5 are the multiples of 5.

These numbers are $\{5, 10, 15, \dots, 90\}$.

To find the number of such numbers, we can divide the last number by 5: $90 \div 5 = 18$.

So, there are 18 numbers divisible by 5 between 1 and 90.

Number of favourable outcomes (divisible by 5) = 18.

The probability of drawing a number divisible by 5 is:

$P(\text{divisible by 5}) = \frac{\text{Number of discs divisible by 5}}{\text{Total number of discs}}$

$P(\text{divisible by 5}) = \frac{18}{90}$

Simplify the fraction by dividing numerator and denominator by 18.

$P(\text{divisible by 5}) = \frac{\cancel{18}^{1}}{\cancel{90}_{5}}$

$P(\text{divisible by 5}) = \frac{1}{5}$

---

Answer:

(i) The probability that it bears a two-digit number is $\mathbf{\frac{9}{10}}$.

(ii) The probability that it bears a perfect square number is $\mathbf{\frac{1}{10}}$.

(iii) The probability that it bears a number divisible by 5 is $\mathbf{\frac{1}{5}}$.

Given:

A die with six faces showing the letters: A, B, C, D, E, A.

The die is thrown once.

---

To Find:

(i) The probability of getting the letter A.

(ii) The probability of getting the letter D.

---

Solution:

The letters on the six faces of the die are A, B, C, D, E, A.

The total number of possible outcomes when the die is thrown once is the number of faces on the die, which is 6.

Total number of possible outcomes = 6.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

---

(i) Probability of getting A:

Let E$_{1}$ be the event of getting the letter A.

The letter A appears on 2 faces of the die.

Number of favourable outcomes for E$_{1}$ (A) = 2.

The probability of getting A is:

$P(E_{1}) = P(\text{A}) = \frac{\text{Number of faces with A}}{\text{Total number of faces}}$

$P(\text{A}) = \frac{2}{6}$

Simplify the fraction by dividing the numerator and denominator by 2.

$P(\text{A}) = \frac{\cancel{2}^{1}}{\cancel{6}_{3}}$

$P(\text{A}) = \frac{1}{3}$

---

(ii) Probability of getting D:

Let E$_{2}$ be the event of getting the letter D.

The letter D appears on 1 face of the die.

Number of favourable outcomes for E$_{2}$ (D) = 1.

The probability of getting D is:

$P(E_{2}) = P(\text{D}) = \frac{\text{Number of faces with D}}{\text{Total number of faces}}$

$P(\text{D}) = \frac{1}{6}$

---

Answer:

(i) The probability of getting A is $\mathbf{\frac{1}{3}}$.

(ii) The probability of getting D is $\mathbf{\frac{1}{6}}$.

Given:

A rectangular region with dimensions 3 m by 2 m.

A circle inside the rectangle with diameter 1 m.

A die is dropped at random on the rectangular region.

---

To Find:

The probability that the die will land inside the circle.

---

Solution:

This is a geometric probability problem, where the probability is determined by the ratio of areas.

The sample space is the entire rectangular region. The area of the rectangular region is the total possible area where the die can land.

Area of rectangle = length $\times$ width

Area of rectangle = $3 \text{ m} \times 2 \text{ m} = 6$ square m.

The favourable event is that the die lands inside the circle. The area of the circle is the favourable area.

The diameter of the circle is 1 m. The radius of the circle is half of the diameter.

Radius of circle, $r = \frac{1}{2}$ m.

Area of circle = $\pi r^2$

Area of circle = $\pi \left(\frac{1}{2}\right)^2 = \pi \times \frac{1}{4} = \frac{\pi}{4}$ square m.

The probability that the die lands inside the circle is the ratio of the area of the circle to the area of the rectangular region.

Probability (lands inside circle) = $\frac{\text{Area of circle}}{\text{Area of rectangle}}$

Probability (lands inside circle) = $\frac{\frac{\pi}{4} \text{ m}^2}{6 \text{ m}^2}$

Probability (lands inside circle) = $\frac{\pi/4}{6}$

To simplify the fraction, we can write:

$\frac{\pi/4}{6} = \frac{\pi}{4} \times \frac{1}{6} = \frac{\pi}{24}$

---

Answer:

The probability that the die will land inside the circle is $\mathbf{\frac{\pi}{24}}$.

Given:

Total number of ball pens in the lot = 144.

Number of defective pens = 20.

---

To Find:

(i) The probability that Nuri will buy the pen.

(ii) The probability that Nuri will not buy the pen.

---

Solution:

First, calculate the number of good pens in the lot.

Number of good pens = Total number of pens - Number of defective pens

Number of good pens = $144 - 20 = 124$.

When the shopkeeper draws one pen at random, the total number of possible outcomes is the total number of pens, which is 144.

Total number of possible outcomes = 144.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

---

(i) Probability that she will buy it:

Nuri will buy a pen if and only if it is good.

The event is drawing a good pen.

The number of favourable outcomes for this event is the number of good pens.

Number of favourable outcomes (good pen) = 124.

The probability that she will buy the pen is:

$P(\text{she will buy}) = P(\text{good pen}) = \frac{\text{Number of good pens}}{\text{Total number of pens}}$

$P(\text{she will buy}) = \frac{124}{144}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4.

$P(\text{she will buy}) = \frac{\cancel{124}^{31}}{\cancel{144}_{36}}$

$P(\text{she will buy}) = \frac{31}{36}$

---

(ii) Probability that she will not buy it:

Nuri will not buy a pen if and only if it is defective.

The event is drawing a defective pen.

The number of favourable outcomes for this event is the number of defective pens.

Number of favourable outcomes (defective pen) = 20.

The probability that she will not buy the pen is:

$P(\text{she will not buy}) = P(\text{defective pen}) = \frac{\text{Number of defective pens}}{\text{Total number of pens}}$

$P(\text{she will not buy}) = \frac{20}{144}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4.

$P(\text{she will not buy}) = \frac{\cancel{20}^{5}}{\cancel{144}_{36}}$

$P(\text{she will not buy}) = \frac{5}{36}$

---

Alternate Solution for (ii):

The event "she will buy the pen" and the event "she will not buy the pen" are complementary events.

Using the complementary event formula: $P(\text{she will not buy}) = 1 - P(\text{she will buy})$.

From part (i), we found $P(\text{she will buy}) = \frac{31}{36}$.

So, $P(\text{she will not buy}) = 1 - \frac{31}{36}$

$P(\text{she will not buy}) = \frac{36}{36} - \frac{31}{36}$

$P(\text{she will not buy}) = \frac{36-31}{36}$

$P(\text{she will not buy}) = \frac{5}{36}$

This confirms the result obtained earlier.

---

Answer:

(i) The probability that she will buy it is $\mathbf{\frac{31}{36}}$.

(ii) The probability that she will not buy it is $\mathbf{\frac{5}{36}}$.

Given:

Two dice, one blue and one grey, are thrown simultaneously. The possible sums of the numbers appearing on the top faces range from 2 to 12.

---

Solution:

The sample space when two dice are thrown consists of 36 equally likely outcomes, as listed in Example 13:

$S = \{$

$(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),$

$(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),$

$(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),$

$(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),$

$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),$

$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$

$\}$

Total number of outcomes = 36.

The number of outcomes that result in a specific sum is the count of pairs from the sample space that add up to that sum.

---

(i) Complete the following table:

Let's find the number of favourable outcomes for each sum:

Sum = 2: (1,1) - 1 outcome. $P(\text{Sum}=2) = \frac{1}{36}$.

Sum = 3: (1,2), (2,1) - 2 outcomes. $P(\text{Sum}=3) = \frac{2}{36} = \frac{1}{18}$.

Sum = 4: (1,3), (2,2), (3,1) - 3 outcomes. $P(\text{Sum}=4) = \frac{3}{36} = \frac{1}{12}$.

Sum = 5: (1,4), (2,3), (3,2), (4,1) - 4 outcomes. $P(\text{Sum}=5) = \frac{4}{36} = \frac{1}{9}$.

Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) - 5 outcomes. $P(\text{Sum}=6) = \frac{5}{36}$.

Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - 6 outcomes. $P(\text{Sum}=7) = \frac{6}{36} = \frac{1}{6}$.

Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) - 5 outcomes. $P(\text{Sum}=8) = \frac{5}{36}$ (given).

Sum = 9: (3,6), (4,5), (5,4), (6,3) - 4 outcomes. $P(\text{Sum}=9) = \frac{4}{36} = \frac{1}{9}$.

Sum = 10: (4,6), (5,5), (6,4) - 3 outcomes. $P(\text{Sum}=10) = \frac{3}{36} = \frac{1}{12}$.

Sum = 11: (5,6), (6,5) - 2 outcomes. $P(\text{Sum}=11) = \frac{2}{36} = \frac{1}{18}$.

Sum = 12: (6,6) - 1 outcome. $P(\text{Sum}=12) = \frac{1}{36}$ (given).

Completed table:

Event: Sum on 2 dice	2	3	4	5	6	7	8	9	10	11	12
Probability	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

---

(ii) Do you agree with this argument? Justify your answer.

The argument is that since there are 11 possible sums (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), each sum has a probability of $\frac{1}{11}$.

I do not agree with this argument.

Justification:

The argument is incorrect because the 11 possible outcomes (the sums) are not equally likely.

For outcomes to be equally likely, each outcome must have the same probability of occurring.

When rolling two dice, the elementary outcomes are the 36 pairs $(x, y)$, where $x$ and $y$ are the numbers on the individual dice. Each of these 36 elementary outcomes has a probability of $\frac{1}{36}$.

The sum of the two numbers is a derived outcome, and different sums can be obtained in a different number of ways (as shown in the table above).

For example, the sum of 2 occurs in only 1 way ((1,1)), so its probability is $\frac{1}{36}$.

The sum of 7 occurs in 6 ways ((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)), so its probability is $\frac{6}{36} = \frac{1}{6}$.

Since $\frac{1}{36} \neq \frac{1}{6}$, the sum outcomes are not equally likely.

Therefore, it is incorrect to assume that each sum has a probability of $\frac{1}{11}$. The probability depends on the number of elementary outcomes that result in that sum.

Given:

A one rupee coin is tossed 3 times.

Hanif wins if the outcome is three heads (HHH) or three tails (TTT).

Hanif loses otherwise.

---

To Find:

The probability that Hanif will lose the game.

---

Solution:

When a coin is tossed 3 times, the possible outcomes for the sequence of results are:

H for Head, T for Tail.

The sample space S is the set of all possible outcomes:

$S = \{\text{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}\}$

The total number of possible outcomes is the number of elements in the sample space.

Total number of outcomes = $|S| = 8$.

Hanif wins if the result is HHH or TTT.

Let W be the event that Hanif wins.

The favourable outcomes for W are $\{$HHH, TTT$\}$.

Number of favourable outcomes for W = 2.

Hanif loses if the outcome is not HHH and not TTT. This is the complementary event to winning.

Let L be the event that Hanif loses.

The favourable outcomes for L are the outcomes in S that are not in W:

$L = \{\text{HHT, HTH, THH, HTT, THT, TTH}\}$

Number of favourable outcomes for L = 6.

The probability of an event is given by the formula:

$P(\text{event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

The probability that Hanif will lose the game is:

$P(\text{Hanif loses}) = \frac{\text{Number of outcomes where Hanif loses}}{\text{Total number of possible outcomes}}$

$P(\text{Hanif loses}) = \frac{6}{8}$

Simplify the fraction by dividing the numerator and denominator by 2:

$P(\text{Hanif loses}) = \frac{\cancel{6}^{3}}{\cancel{8}_{4}}$

$P(\text{Hanif loses}) = \frac{3}{4}$

---

Alternate Solution:

The event that Hanif loses is the complement of the event that Hanif wins.

The probability that Hanif wins is:

$P(\text{Hanif wins}) = \frac{\text{Number of outcomes where Hanif wins}}{\text{Total number of possible outcomes}} = \frac{2}{8} = \frac{1}{4}$

The probability that Hanif loses is $1$ minus the probability that Hanif wins:

$P(\text{Hanif loses}) = 1 - P(\text{Hanif wins})$

$P(\text{Hanif loses}) = 1 - \frac{1}{4}$

$P(\text{Hanif loses}) = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}$

Both methods give the same result.

---

Answer:

The probability that Hanif will lose the game is $\mathbf{\frac{3}{4}}$.

Given:

A die is thrown twice.

---

To Find:

(i) The probability that 5 will not come up either time.

(ii) The probability that 5 will come up at least once.

---

Solution:

When a die is thrown twice, the outcome is an ordered pair $(x, y)$, where $x$ is the result of the first throw and $y$ is the result of the second throw. Both $x$ and $y$ can be any integer from 1 to 6.

The sample space S is the set of all possible outcomes:

$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),$

$(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),$

$(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),$

$(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),$

$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),$

$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

The total number of possible outcomes is $6 \times 6 = 36$.

Total number of possible outcomes = 36.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability that 5 will not come up either time:

Let E$_{1}$ be the event that 5 does not come up on the first throw AND 5 does not come up on the second throw.

For the first throw, the possible outcomes are $\{1, 2, 3, 4, 6\}$. There are 5 such outcomes.

For the second throw, the possible outcomes are $\{1, 2, 3, 4, 6\}$. There are 5 such outcomes.

The number of outcomes where 5 does not come up either time is the product of the number of possibilities for each throw.

Number of favourable outcomes for E$_{1}$ = $5 \times 5 = 25$.

The probability of event E$_{1}$ is:

$P(E_{1}) = P(\text{5 will not come up either time}) = \frac{\text{Number of outcomes with no 5}}{\text{Total number of outcomes}}$

$P(\text{5 will not come up either time}) = \frac{25}{36}$

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(ii) Probability that 5 will come up at least once:

Let E$_{2}$ be the event that 5 will come up at least once. This means 5 comes up on the first throw, or on the second throw, or on both throws.

The outcomes where 5 comes up at least once are the outcomes in the sample space where either the first coordinate is 5 or the second coordinate is 5 (or both).

Outcomes with 5 on the first throw: $\{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$ (6 outcomes).

Outcomes with 5 on the second throw: $\{(1,5), (2,5), (3,5), (4,5), (6,5)\}$ (5 outcomes - note that (5,5) is already counted in the first set).

The set of favourable outcomes for E$_{2}$ is $\{(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6)\}$.

Number of favourable outcomes for E$_{2}$ = $6 + 5 = 11$.

The probability of event E$_{2}$ is:

$P(E_{2}) = P(\text{5 will come up at least once}) = \frac{\text{Number of outcomes with at least one 5}}{\text{Total number of outcomes}}$

$P(\text{5 will come up at least once}) = \frac{11}{36}$

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Alternate Solution for (ii):

The event "5 will come up at least once" is the complementary event to "5 will not come up either time".

Let E be the event "5 will come up at least once".

Let $\bar{E}$ be the event "5 will not come up either time".

Using the complementary event formula: $P(E) + P(\bar{E}) = 1$.

From part (i), we found $P(\bar{E}) = P(\text{5 will not come up either time}) = \frac{25}{36}$.

So, $P(\text{5 will come up at least once}) = 1 - P(\text{5 will not come up either time})$

$P(\text{5 will come up at least once}) = 1 - \frac{25}{36}$

$P(\text{5 will come up at least once}) = \frac{36}{36} - \frac{25}{36}$

$P(\text{5 will come up at least once}) = \frac{36-25}{36}$

$P(\text{5 will come up at least once}) = \frac{11}{36}$

This confirms the result obtained earlier.

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Answer:

(i) The probability that 5 will not come up either time is $\mathbf{\frac{25}{36}}$.

(ii) The probability that 5 will come up at least once is $\mathbf{\frac{11}{36}}$.

(i) Incorrect.

Reason: When two coins are tossed simultaneously, the set of equally likely **elementary outcomes** is:

$S = \{\text{HH, HT, TH, TT}\}$

The total number of equally likely elementary outcomes is 4.

The event "two heads" corresponds to the outcome {HH}, which has 1 favourable outcome.

$P(\text{two heads}) = \frac{1}{4}$

The event "two tails" corresponds to the outcome {TT}, which has 1 favourable outcome.

$P(\text{two tails}) = \frac{1}{4}$

The event "one of each" corresponds to the outcomes {HT, TH}, which has 2 favourable outcomes.

$P(\text{one of each}) = \frac{2}{4} = \frac{1}{2}$

Since the probabilities $\frac{1}{4}$, $\frac{1}{4}$, and $\frac{1}{2}$ are not equal, the three stated outcomes (two heads, two tails, one of each) are not equally likely. Therefore, the assumption that each has a probability of $\frac{1}{3}$ is incorrect.

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(ii) Correct.

Reason: When a standard die is thrown, the set of equally likely **elementary outcomes** is:

$S = \{1, 2, 3, 4, 5, 6\}$

The total number of equally likely elementary outcomes is 6.

The event "getting an odd number" corresponds to the elementary outcomes {1, 3, 5}.

Number of favourable outcomes for 'odd number' = 3.

$P(\text{getting an odd number}) = \frac{\text{Number of odd outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$

The event "getting an even number" corresponds to the elementary outcomes {2, 4, 6}.

Number of favourable outcomes for 'even number' = 3.

$P(\text{getting an even number}) = \frac{\text{Number of even outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$

In this case, although 'odd number' and 'even number' are not elementary outcomes, the set of elementary outcomes is divided into two subsets (odd and even) that contain an equal number of elementary outcomes. Hence, the events 'getting an odd number' and 'getting an even number' are equally likely, each with a probability of $\frac{3}{6} = \frac{1}{2}$. The argument is correct.

Given:

Two customers, Shyam and Ekta, visit a shop between Tuesday and Saturday.

The days are Tuesday, Wednesday, Thursday, Friday, Saturday.

Each customer is equally likely to visit the shop on any of these days.

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To Find:

The probability that both will visit the shop on:

(i) the same day.

(ii) consecutive days.

(iii) different days.

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Solution:

The number of days the shop is visited is 5 (Tuesday to Saturday).

Let the days be represented by numbers 1, 2, 3, 4, 5 for Tuesday, Wednesday, Thursday, Friday, Saturday respectively.

Since each customer can visit on any of the 5 days, the total number of possible pairs of visiting days for Shyam and Ekta is $5 \times 5 = 25$.

The sample space S consists of ordered pairs (Shyam's day, Ekta's day). For example, (Tuesday, Wednesday) means Shyam visits on Tuesday and Ekta visits on Wednesday.

Total number of possible outcomes = 25.

Since each day is equally likely for each customer, each of the 25 outcomes in the sample space is equally likely.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

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(i) Probability that both will visit the shop on the same day:

Let E$_{1}$ be the event that Shyam and Ekta visit on the same day.

The favourable outcomes are when their visiting days are the same:

E$_{1} = \{(\text{Tue, Tue}), (\text{Wed, Wed}), (\text{Thu, Thu}), (\text{Fri, Fri}), (\text{Sat, Sat})\}$

Number of favourable outcomes for E$_{1}$ = 5.

The probability of event E$_{1}$ is:

$P(\text{same day}) = \frac{5}{25}$

Simplify the fraction:

$P(\text{same day}) = \frac{1}{5}$

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(ii) Probability that both will visit the shop on consecutive days:

Let E$_{2}$ be the event that Shyam and Ekta visit on consecutive days.

Consecutive days mean either Ekta visits the day after Shyam or Shyam visits the day after Ekta.

The consecutive pairs of days from Tuesday to Saturday are:

(Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday)

And the reverse pairs:

(Wednesday, Tuesday), (Thursday, Wednesday), (Friday, Thursday), (Saturday, Friday)

The favourable outcomes for E$_{2}$ are:

E$_{2} = \{(\text{Tue, Wed}), (\text{Wed, Thu}), (\text{Thu, Fri}), (\text{Fri, Sat}), (\text{Wed, Tue}), (\text{Thu, Wed}), (\text{Fri, Thu}), (\text{Sat, Fri})\}$

Number of favourable outcomes for E$_{2}$ = 8.

The probability of event E$_{2}$ is:

$P(\text{consecutive days}) = \frac{8}{25}$

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(iii) Probability that both will visit the shop on different days:

Let E$_{3}$ be the event that Shyam and Ekta visit on different days.

The total number of outcomes is 25. The event "different days" is the complement of the event "same day".

Number of favourable outcomes for E$_{3}$ = (Total number of outcomes) - (Number of outcomes for same day)

Number of favourable outcomes for E$_{3}$ = $25 - 5 = 20$.

The probability of event E$_{3}$ is:

$P(\text{different days}) = \frac{20}{25}$

Simplify the fraction:

$P(\text{different days}) = \frac{\cancel{20}^{4}}{\cancel{25}_{5}}$

$P(\text{different days}) = \frac{4}{5}$

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Alternate Solution for (iii):

The event "different days" is the complementary event to "the same day".

Using the complementary event formula: $P(\text{different days}) = 1 - P(\text{same day})$.

From part (i), $P(\text{same day}) = \frac{1}{5}$.

So, $P(\text{different days}) = 1 - \frac{1}{5}$

$P(\text{different days}) = \frac{5}{5} - \frac{1}{5} = \frac{4}{5}$

This confirms the result obtained earlier.

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Answer:

(i) The probability that both will visit the shop on the same day is $\mathbf{\frac{1}{5}}$.

(ii) The probability that both will visit the shop on consecutive days is $\mathbf{\frac{8}{25}}$.

(iii) The probability that both will visit the shop on different days is $\mathbf{\frac{4}{5}}$.

Given:

A die has faces showing the numbers 1, 2, 2, 3, 3, 6. The die is thrown two times.

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Solution:

When the die is thrown twice, there are 6 possible outcomes for the first throw and 6 possible outcomes for the second throw. The total number of equally likely outcomes is $6 \times 6 = 36$. Each outcome is an ordered pair $(f_1, f_2)$ where $f_1$ is the number on the first throw and $f_2$ is the number on the second throw.

The numbers on the faces are {1, 2, 2, 3, 3, 6}. The 36 possible sums are shown in the table provided in the question (Fig. 15.7). We can count the frequency of each sum from this table:

Sum = 2: (1,1) - 1 outcome

Sum = 3: (1,2), (1,2), (2,1), (2,1) - 4 outcomes

Sum = 4: (1,3), (1,3), (2,2), (2,2), (2,2), (2,2), (3,1), (3,1) - 8 outcomes

Sum = 5: (2,3), (2,3), (2,3), (2,3), (3,2), (3,2), (3,2), (3,2) - 8 outcomes

Sum = 6: (3,3), (3,3), (3,3), (3,3) - 4 outcomes

Sum = 7: (1,6), (6,1) - 2 outcomes

Sum = 8: (2,6), (2,6), (6,2), (6,2) - 4 outcomes

Sum = 9: (3,6), (3,6), (6,3), (6,3) - 4 outcomes

Sum = 10: No outcomes

Sum = 11: No outcomes

Sum = 12: (6,6) - 1 outcome

Total number of outcomes = $1+4+8+8+4+2+4+4+0+0+1 = 36$.

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(i) Complete the following table:

The probability of each sum is the number of outcomes for that sum divided by the total number of outcomes (36).

Event: Sum on 2 dice	2	3	4	5	6	7	8	9	10	11	12
Probability	$\frac{1}{36}$	$\frac{4}{36}$	$\frac{8}{36}$	$\frac{8}{36}$	$\frac{4}{36}$	$\frac{2}{36}$	$\frac{4}{36}$	$\frac{4}{36}$	$\frac{0}{36}$	$\frac{0}{36}$	$\frac{1}{36}$

Simplifying the fractions:

Event: Sum on 2 dice	2	3	4	5	6	7	8	9	10	11	12
Probability	$\frac{1}{36}$	$\frac{1}{9}$	$\frac{2}{9}$	$\frac{2}{9}$	$\frac{1}{9}$	$\frac{1}{18}$	$\frac{1}{9}$	$\frac{1}{9}$	0	0	$\frac{1}{36}$

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(ii) Probability that the total score is even:

The even sums are 2, 4, 6, 8, and 12.

Number of outcomes for even sums = (Count for Sum 2) + (Count for Sum 4) + (Count for Sum 6) + (Count for Sum 8) + (Count for Sum 12)

Number of outcomes for even sums = $1 + 8 + 4 + 4 + 1 = 18$.

The probability is:

$P(\text{even score}) = \frac{\text{Number of outcomes with even sum}}{\text{Total number of outcomes}}$

$P(\text{even score}) = \frac{18}{36}$

Simplify the fraction:

$P(\text{even score}) = \frac{\cancel{18}^{1}}{\cancel{36}_{2}}$

$P(\text{even score}) = \frac{1}{2}$

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(iii) Probability that the total score is 6:

The number of outcomes where the sum is 6 is 4.

The probability is:

$P(\text{score is 6}) = \frac{\text{Number of outcomes with sum 6}}{\text{Total number of outcomes}}$

$P(\text{score is 6}) = \frac{4}{36}$

Simplify the fraction:

$P(\text{score is 6}) = \frac{\cancel{4}^{1}}{\cancel{36}_{9}}$

$P(\text{score is 6}) = \frac{1}{9}$

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(iv) Probability that the total score is at least 6:

The sums that are at least 6 are 6, 7, 8, 9, and 12.

Number of outcomes for sums $\geq 6$ = (Count for Sum 6) + (Count for Sum 7) + (Count for Sum 8) + (Count for Sum 9) + (Count for Sum 12)

Number of outcomes for sums $\geq 6$ = $4 + 2 + 4 + 4 + 1 = 15$.

The probability is:

$P(\text{score is at least 6}) = \frac{\text{Number of outcomes with sum} \geq 6}{\text{Total number of outcomes}}$

$P(\text{score is at least 6}) = \frac{15}{36}$

Simplify the fraction:

$P(\text{score is at least 6}) = \frac{\cancel{15}^{5}}{\cancel{36}_{12}}$

$P(\text{score is at least 6}) = \frac{5}{12}$

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Answer:

(i) The completed table is shown above.

(ii) The probability that the total score is even is $\mathbf{\frac{1}{2}}$.

(iii) The probability that the total score is 6 is $\mathbf{\frac{1}{9}}$.

(iv) The probability that the total score is at least 6 is $\mathbf{\frac{5}{12}}$.

Given:

Number of red balls = 5.

Let the number of blue balls be B.

The probability of drawing a blue ball is double that of drawing a red ball, which can be written as $P(\text{Blue}) = 2 \times P(\text{Red})$.

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To Find:

The number of blue balls in the bag (B).

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Solution:

The total number of balls in the bag is the sum of the number of red balls and the number of blue balls.

Total number of balls = $5 + B$.

The probability of drawing a red ball is given by:

$P(\text{Red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{5}{5+B}$

The probability of drawing a blue ball is given by:

$P(\text{Blue}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{B}{5+B}$

We are given the condition:

$P(\text{Blue}) = 2 \times P(\text{Red})$

Substitute the expressions for $P(\text{Blue})$ and $P(\text{Red})$ into the equation:

$\frac{B}{5+B} = 2 \times \frac{5}{5+B}$

Simplify the right side of the equation:

$\frac{B}{5+B} = \frac{10}{5+B}$

Since B represents the number of blue balls, $B \geq 0$. Therefore, the total number of balls $5+B$ is greater than or equal to 5, and thus $5+B \neq 0$.

We can multiply both sides of the equation by $(5+B)$:

$(5+B) \times \frac{B}{5+B} = (5+B) \times \frac{10}{5+B}$

$B = 10$

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Check:

If B = 10, then total balls = $5 + 10 = 15$.

$P(\text{Red}) = \frac{5}{15} = \frac{1}{3}$.

$P(\text{Blue}) = \frac{10}{15} = \frac{2}{3}$.

Is $P(\text{Blue}) = 2 \times P(\text{Red})$?

$\frac{2}{3} = 2 \times \frac{1}{3} = \frac{2}{3}$. Yes, the condition is satisfied.

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Answer:

The number of blue balls in the bag is $\mathbf{10}$.

Given:

Initial number of balls in the box = 12.

Initial number of black balls = x.

6 more black balls are added to the box.

The new probability of drawing a black ball is double the initial probability.

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To Find:

The probability of drawing a black ball initially (in terms of x).

The value of x (the initial number of black balls).

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Solution:

Initially, the box contains 12 balls, and x of them are black.

Total number of possible outcomes when drawing one ball initially = 12.

The number of favourable outcomes for drawing a black ball initially = x.

The probability of drawing a black ball initially, $P_1(\text{Black})$, is:

$P_1(\text{Black}) = \frac{\text{Number of black balls}}{\text{Total number of balls}}$

$P_1(\text{Black}) = \frac{x}{12}$

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Now, 6 more black balls are put into the box.

New number of black balls = $x + 6$.

New total number of balls = Initial total balls + Added black balls

New total number of balls = $12 + 6 = 18$.

The probability of drawing a black ball from the new box, $P_2(\text{Black})$, is:

$P_2(\text{Black}) = \frac{\text{New number of black balls}}{\text{New total number of balls}}$

$P_2(\text{Black}) = \frac{x+6}{18}$

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According to the problem statement, the new probability of drawing a black ball is double the initial probability.

$P_2(\text{Black}) = 2 \times P_1(\text{Black})$

Substitute the expressions for the probabilities:

Simplify the equation:

$\frac{x+6}{18} = \frac{2x}{12}$

Simplify the fraction on the right side:

$\frac{x+6}{18} = \frac{x}{6}$

Multiply both sides by the least common multiple of 18 and 6, which is 18, to clear the denominators:

$18 \times \frac{x+6}{18} = 18 \times \frac{x}{6}$

$x+6 = 3x$

Subtract x from both sides:

$6 = 3x - x$

$6 = 2x$

Divide by 2:

$x = \frac{6}{2}$

$x = 3$

Since x represents the number of black balls, it must be a non-negative integer. Also, initially, there were 12 balls, so x must be less than or equal to 12. The value $x=3$ satisfies these conditions.

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Answer:

The probability that it will be a black ball initially is $\mathbf{\frac{x}{12}}$.

The value of x is $\mathbf{3}$.

Given:

Total number of marbles in the jar = 24.

The marbles are either green or blue.

The probability of drawing a green marble, $P(\text{Green}) = \frac{2}{3}$.

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To Find:

The number of blue marbles in the jar.

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Solution:

Let the number of green marbles be $G$.

Let the number of blue marbles be $B$.

The total number of marbles in the jar is the sum of green and blue marbles:

$G + B = 24$

The probability of drawing a green marble is given by:

$P(\text{Green}) = \frac{\text{Number of green marbles}}{\text{Total number of marbles}}$

$P(\text{Green}) = \frac{G}{24}$

We are given that the probability of drawing a green marble is $\frac{2}{3}$.

So, we have the equation:

$\frac{G}{24} = \frac{2}{3}$

To find the number of green marbles ($G$), multiply both sides of the equation by 24:

$G = \frac{2}{3} \times 24$

$G = 2 \times \frac{\cancel{24}^{8}}{\cancel{3}_{1}}$

$G = 2 \times 8$

$G = 16$

So, there are 16 green marbles in the jar.

Now, using the total number of marbles:

Number of green marbles + Number of blue marbles = Total marbles

$16 + B = 24$

Subtract 16 from both sides to find the number of blue marbles:

$B = 24 - 16$

$B = 8$

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Answer:

The number of blue marbles in the jar is $\mathbf{8}$.