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Chapter 15 Probability
Welcome to the solutions for Chapter 15: Probability. This chapter introduces the fascinating world of chance and uncertainty, primarily focusing on probability derived from observation and experimentation. While earlier studies might have touched upon theoretical ideas, this chapter specifically delves into experimental or empirical probability. This approach defines probability based on the frequency with which an event occurs during actual trials or experiments. It's a practical perspective grounded in collected data, moving beyond purely theoretical assumptions about fairness or equally likely outcomes, which are often ideals not perfectly reflected in reality. We will explore how to quantify likelihood based on what has actually been observed.
The fundamental concept underpinning this chapter is the calculation of empirical probability. For any specific outcome or set of outcomes we are interested in, referred to as an event (let's denote it as $E$), its empirical probability, written as $P(E)$, is determined by the ratio of the number of times the event actually occurred during an experiment to the total number of times the experiment was conducted (the total number of trials). The formula is precisely expressed as:
$P(E) = \frac{\text{Number of trials in which the event } E \text{ happened}}{\text{Total number of trials}}$
It's crucial to understand the bounds of probability values. As emphasized in the solutions, the probability of any event $E$ must lie within the range of 0 to 1, inclusive. A probability of 0 signifies an impossible event – an event that never occurred in the trials and is considered impossible based on the experiment. Conversely, a probability of 1 represents a sure event – an event that occurred in every single trial conducted. Most events will have probabilities falling somewhere between these two extremes, representing varying degrees of likelihood based on the observed frequencies.
The solutions guide learners through a variety of scenarios where calculating empirical probability from given data is required. These examples serve to solidify the understanding of the core formula and its application:
- Coin Tossing Experiments: Calculating $P(\text{Head})$ or $P(\text{Tail})$ based on the recorded outcomes from, say, $500$ or $1000$ coin flips. For example, if a coin is tossed $\bcancel{||||}$ $\bcancel{||||}$ times (10 times) and heads appear $||||$ times (4 times), then $P(\text{Head}) = \frac{4}{10} = 0.4$.
- Dice Rolling Experiments: Determining the empirical probability of rolling a specific number (e.g., a '5') or a type of number (e.g., 'an even number') based on the results logged from numerous dice rolls.
- Surveys and Polls: Calculating probabilities related to public opinion, preferences, or demographic characteristics derived from survey data. For instance, finding the probability that a randomly selected person from the surveyed group prefers tea over coffee.
- Real-world Observational Data: Analyzing existing records, such as meteorological data to find the probability of rain on a given day based on past records, hospital records for birth statistics, or traffic data for accident frequencies, to calculate empirical probabilities of specific occurrences.
In tackling these problems, the solutions emphasize a clear methodology: first, precisely identify the event of interest ($E$). Second, carefully count the number of trials where this specific event occurred (favorable trials) from the provided data (often presented in tables or descriptive text). Third, identify the total number of trials performed in the experiment or survey. Finally, substitute these counts into the probability formula $P(E) = \frac{\text{Favorable Trials}}{\text{Total Trials}}$ and compute the resulting ratio, often expressed as a fraction or decimal. This chapter provides a robust, data-driven foundation for understanding and quantifying probability based on observed evidence.
Example 1 to 10 (Before Exercise 15.1)
Example 1. A coin is tossed 1000 times with the following frequencies:
Head : 455, Tail : 545
Compute the probability for each event.
Answer:
Given:
Total number of times the coin is tossed = 1000.
Frequency of Heads = 455.
Frequency of Tails = 545.
To Compute:
The probability for each event (getting a Head and getting a Tail).
Solution:
The empirical probability of an event E is given by the formula:
$P(E) = \frac{\text{Number of trials where the event happened}}{\text{Total number of trials}}$
Let H be the event of getting a Head.
Number of times Head appears = 455.
Total number of trials = 1000.
$P(\text{Head}) = P(H) = \frac{\text{Number of Heads}}{\text{Total number of trials}} = \frac{455}{1000}$
$P(\text{Head}) = 0.455$
Let T be the event of getting a Tail.
Number of times Tail appears = 545.
Total number of trials = 1000.
$P(\text{Tail}) = P(T) = \frac{\text{Number of Tails}}{\text{Total number of trials}} = \frac{545}{1000}$
$P(\text{Tail}) = 0.545$
The probability of getting a Head is 0.455.
The probability of getting a Tail is 0.545.
Note that the sum of the probabilities of all possible outcomes is 1:
$P(H) + P(T) = 0.455 + 0.545 = 1.000$
Example 2. Two coins are tossed simultaneously 500 times, and we get
Two heads : 105 times
One head : 275 times
No head : 120 times
Find the probability of occurrence of each of these events.
Answer:
Given:
Total number of times two coins are tossed simultaneously = 500.
Frequency of Two heads = 105.
Frequency of One head = 275.
Frequency of No head = 120.
Let's check the total number of outcomes: $105 + 275 + 120 = 380 + 120 = 500$. This matches the total number of trials.
To Find:
The probability of occurrence of each of these events (getting two heads, getting one head, getting no head).
Solution:
The empirical probability of an event E is given by the formula:
$P(E) = \frac{\text{Number of trials where the event happened}}{\text{Total number of trials}}$
Let E1 be the event of getting two heads.
Number of times E1 happened = 105.
Total number of trials = 500.
$P(\text{Two heads}) = P(E1) = \frac{\text{Number of times Two heads appeared}}{\text{Total number of trials}} = \frac{105}{500}$
$P(\text{Two heads}) = \frac{\cancel{105}^{21}}{\cancel{500}_{100}} = \frac{21}{100} = 0.21$
Let E2 be the event of getting one head.
Number of times E2 happened = 275.
Total number of trials = 500.
$P(\text{One head}) = P(E2) = \frac{\text{Number of times One head appeared}}{\text{Total number of trials}} = \frac{275}{500}$
$P(\text{One head}) = \frac{\cancel{275}^{55}}{\cancel{500}_{100}} = \frac{55}{100} = 0.55$
Let E3 be the event of getting no head.
Number of times E3 happened = 120.
Total number of trials = 500.
$P(\text{No head}) = P(E3) = \frac{\text{Number of times No head appeared}}{\text{Total number of trials}} = \frac{120}{500}$
$P(\text{No head}) = \frac{\cancel{120}^{12}}{\cancel{500}_{50}} = \frac{\cancel{12}^{6}}{\cancel{50}_{25}} = \frac{6}{25} = 0.24$
The probability of getting two heads is 0.21.
The probability of getting one head is 0.55.
The probability of getting no head is 0.24.
Check: Sum of probabilities = $0.21 + 0.55 + 0.24 = 0.76 + 0.24 = 1.00$.
Example 3. A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the following table :
Table 15.6
| Outcome | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 179 | 150 | 157 | 149 | 175 | 190 |
Find the probability of getting each outcome.
Answer:
Given:
A die is thrown 1000 times. The frequencies of each outcome are given in the table:
| Outcome | Frequency |
| 1 | 179 |
| 2 | 150 |
| 3 | 157 |
| 4 | 149 |
| 5 | 175 |
| 6 | 190 |
| Total | 1000 |
Total number of trials = 1000.
To Find:
The probability of getting each outcome (1, 2, 3, 4, 5, and 6).
Solution:
The empirical probability of an event E is calculated as:
$P(E) = \frac{\text{Frequency of the event}}{\text{Total number of trials}}$
Probability of getting 1: $P(1) = \frac{\text{Frequency of 1}}{\text{Total number of trials}} = \frac{179}{1000} = 0.179$
Probability of getting 2: $P(2) = \frac{\text{Frequency of 2}}{\text{Total number of trials}} = \frac{150}{1000} = 0.150$
Probability of getting 3: $P(3) = \frac{\text{Frequency of 3}}{\text{Total number of trials}} = \frac{157}{1000} = 0.157$
Probability of getting 4: $P(4) = \frac{\text{Frequency of 4}}{\text{Total number of trials}} = \frac{149}{1000} = 0.149$
Probability of getting 5: $P(5) = \frac{\text{Frequency of 5}}{\text{Total number of trials}} = \frac{175}{1000} = 0.175$
Probability of getting 6: $P(6) = \frac{\text{Frequency of 6}}{\text{Total number of trials}} = \frac{190}{1000} = 0.190$
The probabilities for each outcome are:
- P(1) = 0.179
- P(2) = 0.150
- P(3) = 0.157
- P(4) = 0.149
- P(5) = 0.175
- P(6) = 0.190
Check: Sum of probabilities = $0.179 + 0.150 + 0.157 + 0.149 + 0.175 + 0.190 = 1.000$.
Example 4. On one page of a telephone directory, there were 200 telephone numbers. The frequency distribution of their unit place digit (for example, in the number 25828573, the unit place digit is 3) is given in Table 15.7 :
Table 15.7
| Digit | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Frequency | 22 | 26 | 22 | 22 | 20 | 10 | 14 | 28 | 16 | 20 |
Without looking at the page, the pencil is placed on one of these numbers, i.e., the number is chosen at random. What is the probability that the digit in its unit place is 6?
Answer:
Given:
Total number of telephone numbers on a page = 200.
The frequency distribution of the unit place digit is given:
| Digit | Frequency |
| 0 | 22 |
| 1 | 26 |
| 2 | 22 |
| 3 | 22 |
| 4 | 20 |
| 5 | 10 |
| 6 | 14 |
| 7 | 28 |
| 8 | 16 |
| 9 | 20 |
| Total | 200 |
Total number of trials (telephone numbers) = 200.
To Find:
The probability that the digit in the unit place of a randomly chosen number is 6.
Solution:
Let E be the event that the digit in the unit place is 6.
From the frequency distribution table, the number of times the digit 6 appears in the unit place is the frequency of the digit 6, which is 14.
Number of times the event E happened (unit digit is 6) = 14.
Total number of trials = 200.
The empirical probability of event E is given by:
$P(E) = \frac{\text{Number of times the unit digit is 6}}{\text{Total number of telephone numbers}}$
$P(\text{Unit digit is 6}) = \frac{14}{200}$
$P(\text{Unit digit is 6}) = \frac{\cancel{14}^{7}}{\cancel{200}_{100}} = \frac{7}{100} = 0.07$
The probability that the digit in the unit place is 6 is 0.07 or $\frac{7}{100}$.
Example 5. The record of a weather station shows that out of the past 250 consecutive days, its weather forecasts were correct 175 times.
(i) What is the probability that on a given day it was correct?
(ii) What is the probability that it was not correct on a given day?
Answer:
Given:
Total number of consecutive days observed = 250.
Number of times the weather forecast was correct = 175.
(i) To Find:
The probability that the forecast was correct on a given day.
Solution (i):
Let E be the event that the weather forecast was correct on a given day.
Number of trials where the event happened (forecast was correct) = 175.
Total number of trials (total days observed) = 250.
The empirical probability of event E is:
$P(E) = \frac{\text{Number of times forecast was correct}}{\text{Total number of days}}$
$P(\text{Forecast correct}) = \frac{175}{250}$
$P(\text{Forecast correct}) = \frac{\cancel{175}^{35}}{\cancel{250}_{50}} = \frac{\cancel{35}^{7}}{\cancel{50}_{10}} = \frac{7}{10} = 0.7$
The probability that on a given day the weather forecast was correct is 0.7 or $\frac{7}{10}$.
(ii) To Find:
The probability that the forecast was not correct on a given day.
Solution (ii):
Let F be the event that the weather forecast was not correct on a given day.
The number of times the forecast was not correct can be found by subtracting the number of correct forecasts from the total number of days.
Number of times forecast was not correct = Total number of days - Number of times forecast was correct
Number of times forecast was not correct = $250 - 175 = 75$
Total number of trials = 250.
The empirical probability of event F is:
$P(F) = \frac{\text{Number of times forecast was not correct}}{\text{Total number of days}}$
$P(\text{Forecast not correct}) = \frac{75}{250}$
$P(\text{Forecast not correct}) = \frac{\cancel{75}^{15}}{\cancel{250}_{50}} = \frac{\cancel{15}^{3}}{\cancel{50}_{10}} = \frac{3}{10} = 0.3$
Alternatively, the event "forecast not correct" is the complement of the event "forecast correct". The sum of the probabilities of an event and its complement is 1.
$P(\text{Forecast not correct}) = 1 - P(\text{Forecast correct})$
$P(\text{Forecast not correct}) = 1 - 0.7 = 0.3$
The probability that on a given day the weather forecast was not correct is 0.3 or $\frac{3}{10}$.
Example 6. A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table shows the results of 1000 cases.
Table 15.8
| Distance (in km) | less than 4000 | 4000 to 9000 | 9001 to 14000 | more than 14000 |
| Frequency | 20 | 210 | 325 | 445 |
If you buy a tyre of this company, what is the probability that :
(i) it will need to be replaced before it has covered 4000 km?
(ii) it will last more than 9000 km?
(iii) it will need to be replaced after it has covered somewhere between 4000 km and 14000 km?
Answer:
Given:
Results of 1000 cases of tyre replacements, based on the distance covered.
| Distance (in km) | Frequency (Number of cases) |
| less than 4000 | 20 |
| 4000 to 9000 | 210 |
| 9001 to 14000 | 325 |
| more than 14000 | 445 |
| Total | 1000 |
Total number of trials (cases) = 1000.
(i) To Find:
The probability that a tyre needs replacement before covering 4000 km.
Solution (i):
Let E1 be the event that a tyre needs to be replaced before covering 4000 km.
From the table, the number of cases where the distance covered is less than 4000 km is 20.
Number of times E1 happened = 20.
Total number of trials = 1000.
The probability of event E1 is:
$P(\text{Distance < 4000 km}) = \frac{\text{Number of cases < 4000 km}}{\text{Total number of cases}}$
$P(< 4000 \text{ km}) = \frac{20}{1000} = \frac{2}{100} = 0.02$
The probability that a tyre will need to be replaced before covering 4000 km is 0.02.
(ii) To Find:
The probability that a tyre will last more than 9000 km.
Solution (ii):
Let E2 be the event that a tyre lasts more than 9000 km.
This includes the cases where the distance covered is from 9001 to 14000 km and more than 14000 km.
Number of cases for 9001 to 14000 km = 325.
Number of cases for more than 14000 km = 445.
Number of times E2 happened = (Cases 9001 to 14000 km) + (Cases more than 14000 km)
Number of times E2 happened = $325 + 445 = 770$
Total number of trials = 1000.
The probability of event E2 is:
$P(\text{Distance > 9000 km}) = \frac{\text{Number of cases > 9000 km}}{\text{Total number of cases}}$
$P(> 9000 \text{ km}) = \frac{770}{1000} = \frac{77}{100} = 0.77$
The probability that a tyre will last more than 9000 km is 0.77.
(iii) To Find:
The probability that a tyre needs replacement after covering between 4000 km and 14000 km (inclusive of 4000 but perhaps exclusive of 14000 based on the class boundaries in the table).
The class intervals are "4000 to 9000" and "9001 to 14000". The wording "between 4000 km and 14000 km" likely refers to the union of these two specific intervals in the table.
Solution (iii):
Let E3 be the event that a tyre needs to be replaced after covering somewhere between 4000 km and 14000 km.
This corresponds to the cases in the intervals "4000 to 9000" and "9001 to 14000".
Number of cases for 4000 to 9000 km = 210.
Number of cases for 9001 to 14000 km = 325.
Number of times E3 happened = (Cases 4000 to 9000 km) + (Cases 9001 to 14000 km)
Number of times E3 happened = $210 + 325 = 535$
Total number of trials = 1000.
The probability of event E3 is:
$P(\text{4000 km} \leq \text{Distance} \leq \text{14000 km}) = \frac{\text{Number of cases between 4000 and 14000 km}}{\text{Total number of cases}}$
$P(\text{4000 to 14000 km}) = \frac{535}{1000} = 0.535$
The probability that a tyre will need to be replaced after it has covered somewhere between 4000 km and 14000 km is 0.535.
Example 7. The percentage of marks obtained by a student in the monthly unit tests are given below:
Table 15.9
| Unit test | I | II | III | IV | V |
| Percentage of marks obtained | 69 | 71 | 73 | 68 | 74 |
Based on this data, find the probability that the student gets more than 70% marks in a unit test.
Answer:
Given:
The percentage of marks obtained by a student in 5 unit tests:
Test I: 69%
Test II: 71%
Test III: 73%
Test IV: 68%
Test V: 74%
Total number of unit tests (trials) = 5.
To Find:
The probability that the student gets more than 70% marks in a unit test.
Solution:
Let E be the event that the student gets more than 70% marks in a unit test.
We need to count how many times the student scored more than 70% marks from the given data:
- Test I: 69% (Not more than 70%)
- Test II: 71% (More than 70%)
- Test III: 73% (More than 70%)
- Test IV: 68% (Not more than 70%)
- Test V: 74% (More than 70%)
The student scored more than 70% marks in 3 tests (Test II, Test III, Test V).
Number of times event E happened (scored > 70%) = 3.
Total number of trials = 5.
The empirical probability of event E is:
$P(E) = \frac{\text{Number of times marks > 70%}}{\text{Total number of tests}}$
$P(\text{Marks > 70%}) = \frac{3}{5}$
$P(\text{Marks > 70%}) = 0.6$
The probability that the student gets more than 70% marks in a unit test is 0.6 or $\frac{3}{5}$.
Example 8. An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained are given in the following table:
Table 15.10
| Age of drivers (in years) | Accidents in one year | ||||
| 0 | 1 | 2 | 3 | over 3 | |
| 18 - 29 | 440 | 160 | 110 | 61 | 35 |
| 30 - 50 | 505 | 125 | 60 | 22 | 18 |
| Above 50 | 360 | 45 | 35 | 15 | 9 |
Find the probabilities of the following events for a driver chosen at random from the city:
(i) being 18-29 years of age and having exactly 3 accidents in one year.
(ii) being 30-50 years of age and having one or more accidents in a year.
(iii) having no accidents in one year.
Answer:
Given:
Data from a survey of 2000 drivers relating age and accidents in one year.
| Age of drivers (in years) | Accidents in one year | ||||
| 0 | 1 | 2 | 3 | over 3 | |
| 18 - 29 | 440 | 160 | 110 | 61 | 35 |
| 30 - 50 | 505 | 125 | 60 | 22 | 18 |
| Above 50 | 360 | 45 | 35 | 15 | 9 |
| Total | 1305 | 330 | 205 | 98 | 62 |
Total number of drivers surveyed = 2000.
To Find:
The probabilities of the given events for a randomly chosen driver.
Solution:
The empirical probability of an event E is $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$.
Total number of trials = 2000 (total number of drivers).
(i) Probability of being 18-29 years of age and having exactly 3 accidents in one year.
Let E1 be the event that a driver is 18-29 years old and has exactly 3 accidents.
From the table, locate the row for the age group "18 - 29" and the column for "Accidents in one year = 3". The number of drivers in this category is 61.
Number of favourable outcomes for E1 = 61.
$P(E1) = \frac{\text{Number of drivers aged 18-29 with 3 accidents}}{\text{Total number of drivers}}$
$P(\text{18-29 years and 3 accidents}) = \frac{61}{2000}$
$P(\text{18-29 years and 3 accidents}) = 0.0305$
The probability is 0.0305.
(ii) Probability of being 30-50 years of age and having one or more accidents in a year.
Let E2 be the event that a driver is 30-50 years old and has one or more accidents.
From the table, locate the row for the age group "30 - 50". We need the number of drivers in this age group with 1 accident, 2 accidents, 3 accidents, or over 3 accidents.
Number of drivers aged 30-50 with 1 accident = 125.
Number of drivers aged 30-50 with 2 accidents = 60.
Number of drivers aged 30-50 with 3 accidents = 22.
Number of drivers aged 30-50 with over 3 accidents = 18.
Number of favourable outcomes for E2 = $125 + 60 + 22 + 18 = 225$
$P(E2) = \frac{\text{Number of drivers aged 30-50 with $\geq$ 1 accident}}{\text{Total number of drivers}}$
$P(\text{30-50 years and $\geq$ 1 accident}) = \frac{225}{2000}$
$P(\text{30-50 years and $\geq$ 1 accident}) = \frac{\cancel{225}^{45}}{\cancel{2000}_{400}} = \frac{\cancel{45}^{9}}{\cancel{400}_{80}} = \frac{9}{80} = 0.1125$
The probability is 0.1125.
(iii) Probability of having no accidents in one year.
Let E3 be the event that a driver has no accidents in one year.
From the table, we need the total number of drivers who had 0 accidents, across all age groups.
Number of drivers with 0 accidents (18-29 years) = 440.
Number of drivers with 0 accidents (30-50 years) = 505.
Number of drivers with 0 accidents (Above 50 years) = 360.
Number of favourable outcomes for E3 = $440 + 505 + 360 = 1305$
$P(E3) = \frac{\text{Total number of drivers with 0 accidents}}{\text{Total number of drivers}}$
$P(\text{No accidents}) = \frac{1305}{2000}$
$P(\text{No accidents}) = \frac{\cancel{1305}^{261}}{\cancel{2000}_{400}} = 0.6525$
The probability is 0.6525.
Example 9. Consider the frequency distribution table (Table 14.3, Example 4, Chapter 14), which gives the weights of 38 students of a class.
(i) Find the probability that the weight of a student in the class lies in the interval 46-50 kg.
(ii) Give two events in this context, one having probability 0 and the other having probability 1.
Answer:
Given:
The frequency distribution table of weights of 38 students (Table 14.3 from Example 4, Chapter 14):
| Weights (in kg) | Number of students (Frequency) |
| 31 - 35 | 9 |
| 36 - 40 | 5 |
| 41 - 45 | 14 |
| 46 - 50 | 3 |
| 51 - 55 | 1 |
| 56 - 60 | 2 |
| 61 - 65 | 2 |
| 66 - 70 | 1 |
| 71 - 75 | 1 |
| Total | 38 |
Total number of students = 38.
(i) To Find:
The probability that the weight of a student lies in the interval 46-50 kg.
Solution (i):
Let E be the event that the weight of a student is in the interval 46-50 kg.
From the table, the number of students whose weight is in the interval 46-50 kg is 3.
Number of favourable outcomes for E = 3.
Total number of students = 38.
The empirical probability of event E is:
$P(E) = \frac{\text{Number of students in 46-50 kg interval}}{\text{Total number of students}}$
$P(\text{Weight in 46-50 kg}) = \frac{3}{38}$
The probability that the weight of a student lies in the interval 46-50 kg is $\frac{3}{38}$.
(ii) To Give:
Two events in this context, one having probability 0 and the other having probability 1.
Solution (ii):
An event with probability 0 is an impossible event.
An event with probability 1 is a sure or certain event.
Event with probability 0 (Impossible Event):
Consider the event that a student's weight is less than 31 kg from this class. Looking at the table, the minimum weight recorded is in the interval 31-35 kg. No student has a weight recorded as less than 31 kg.
Event: A student's weight is 30 kg.
The lowest weight interval is 31-35 kg. Assuming weights are recorded within these intervals, a weight of 30 kg is not possible based on this data.
Number of students with weight 30 kg = 0.
Probability (Weight = 30 kg) = $\frac{0}{38} = 0$.
So, an event with probability 0 is: A student's weight is less than 31 kg.
Event with probability 1 (Sure Event):
Consider the event that a student's weight is within the range covered by the data. The weights are distributed from 31 kg up to 75 kg (or slightly higher depending on how the last interval is interpreted, but definitely not exceeding the upper limit of the last interval).
Event: A student's weight is between 31 kg and 75 kg (inclusive of 31, possibly inclusive/exclusive of 75 based on original boundaries).
More simply, consider the event that a student's weight falls into one of the given class intervals.
Event: A student's weight is between 31 kg and 75 kg (inclusive). The data covers all weights from 31-35 up to 71-75. Every student in the class falls into one of these categories.
Number of students with weight between 31 kg and 75 kg (inclusive of boundaries) = 38.
Probability (Weight between 31 kg and 75 kg) = $\frac{38}{38} = 1$.
So, an event with probability 1 is: A student's weight is between 31 kg and 75 kg (inclusive).
Example 10. Fifty seeds were selected at random from each of 5 bags of seeds, and were kept under standardised conditions favourable to germination. After 20 days, the number of seeds which had germinated in each collection were counted and recorded as follows:
Table 15.11
| Bag | 1 | 2 | 3 | 4 | 5 |
| Number of seeds germinated | 40 | 48 | 42 | 39 | 41 |
What is the probability of germination of
(i) more than 40 seeds in a bag?
(ii) 49 seeds in a bag?
(iii) more that 35 seeds in a bag?
Answer:
Given:
Fifty seeds were selected from each of 5 bags. The number of seeds germinated after 20 days are:
Bag 1: 40
Bag 2: 48
Bag 3: 42
Bag 4: 39
Bag 5: 41
Total number of bags (trials) = 5.
To Find:
The probability of germination for different scenarios.
Solution:
The empirical probability of an event E is $P(E) = \frac{\text{Number of trials where the event happened}}{\text{Total number of trials}}$.
Total number of trials = 5 (number of bags).
(i) Probability of germination of more than 40 seeds in a bag.
Let E1 be the event that more than 40 seeds germinated in a bag.
We check which bags had more than 40 seeds germinated:
- Bag 1: 40 (Not more than 40)
- Bag 2: 48 (More than 40)
- Bag 3: 42 (More than 40)
- Bag 4: 39 (Not more than 40)
- Bag 5: 41 (More than 40)
The number of bags where more than 40 seeds germinated is 3 (Bag 2, Bag 3, Bag 5).
Number of favourable outcomes for E1 = 3.
$P(E1) = \frac{\text{Number of bags with > 40 seeds germinated}}{\text{Total number of bags}}$
$P(\text{> 40 seeds}) = \frac{3}{5} = 0.6$
The probability of germination of more than 40 seeds in a bag is 0.6 or $\frac{3}{5}$.
(ii) Probability of germination of 49 seeds in a bag.
Let E2 be the event that exactly 49 seeds germinated in a bag.
We check the number of seeds germinated in each bag:
- Bag 1: 40
- Bag 2: 48
- Bag 3: 42
- Bag 4: 39
- Bag 5: 41
None of the bags had exactly 49 seeds germinated.
Number of favourable outcomes for E2 = 0.
$P(E2) = \frac{\text{Number of bags with 49 seeds germinated}}{\text{Total number of bags}}$
$P(\text{49 seeds}) = \frac{0}{5} = 0$
The probability of germination of 49 seeds in a bag is 0.
(iii) Probability of germination of more than 35 seeds in a bag.
Let E3 be the event that more than 35 seeds germinated in a bag.
We check which bags had more than 35 seeds germinated:
- Bag 1: 40 (More than 35)
- Bag 2: 48 (More than 35)
- Bag 3: 42 (More than 35)
- Bag 4: 39 (More than 35)
- Bag 5: 41 (More than 35)
All 5 bags had more than 35 seeds germinated.
Number of favourable outcomes for E3 = 5.
$P(E3) = \frac{\text{Number of bags with > 35 seeds germinated}}{\text{Total number of bags}}$
$P(\text{> 35 seeds}) = \frac{5}{5} = 1$
The probability of germination of more than 35 seeds in a bag is 1.
Exercise 15.1
Question 1. In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Answer:
Given:
Total number of balls played by the batswoman = 30.
Number of times she hits a boundary = 6.
To Find:
The probability that she did not hit a boundary.
Solution:
Total number of trials = 30.
Number of times the event "hitting a boundary" happened = 6.
The number of times the event "not hitting a boundary" happened can be found by subtracting the number of boundaries from the total number of balls.
Number of times she did not hit a boundary = Total balls played - Number of boundaries hit
Number of times she did not hit a boundary = $30 - 6 = 24$
Let E be the event that she did not hit a boundary.
Number of favourable outcomes for E = 24.
Total number of trials = 30.
The empirical probability of event E is:
$P(E) = \frac{\text{Number of times she did not hit a boundary}}{\text{Total number of balls played}}$
$P(\text{Did not hit a boundary}) = \frac{24}{30}$
$P(\text{Did not hit a boundary}) = \frac{\cancel{24}^{4}}{\cancel{30}_{5}} = \frac{4}{5}$
$P(\text{Did not hit a boundary}) = 0.8$
The probability that she did not hit a boundary is $\frac{4}{5}$ or 0.8.
Alternate Solution:
First, find the probability of hitting a boundary.
$P(\text{Hitting a boundary}) = \frac{\text{Number of boundaries}}{\text{Total balls}} = \frac{6}{30} = \frac{1}{5}$
The event "not hitting a boundary" is the complement of the event "hitting a boundary".
$P(\text{Did not hit a boundary}) = 1 - P(\text{Hitting a boundary})$
$P(\text{Did not hit a boundary}) = 1 - \frac{1}{5} = \frac{5}{5} - \frac{1}{5} = \frac{5-1}{5} = \frac{4}{5}$
This gives the same result.
Question 2. 1500 families with 2 children were selected randomly, and the following data were recorded:
| Number of girls in a family | 2 | 1 | 0 |
| Number of families | 475 | 814 | 211 |
Compute the probability of a family, chosen at random, having
(i) 2 girls
(ii) 1 girl
(iii) No girl
Also check whether the sum of these probabilities is 1.
Answer:
Given:
Data from a survey of 1500 families with 2 children.
| Number of girls in a family | Number of families (Frequency) |
| 2 | 475 |
| 1 | 814 |
| 0 | 211 |
| Total | 1500 |
Total number of families surveyed = 1500.
To Compute:
The probability of a randomly chosen family having 2 girls, 1 girl, or no girl.
Solution:
The empirical probability of an event E is $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$.
Total number of trials = 1500 (total number of families).
(i) Probability of a family having 2 girls.
Let E1 be the event that a family has 2 girls.
Number of families with 2 girls = 475.
$P(E1) = \frac{\text{Number of families with 2 girls}}{\text{Total number of families}}$
$P(\text{2 girls}) = \frac{475}{1500}$
$P(\text{2 girls}) = \frac{\cancel{475}^{95}}{\cancel{1500}_{300}} = \frac{\cancel{95}^{19}}{\cancel{300}_{60}} = \frac{19}{60}$
$P(\text{2 girls}) \approx 0.3167$
The probability of a family having 2 girls is $\frac{19}{60}$.
(ii) Probability of a family having 1 girl.
Let E2 be the event that a family has 1 girl.
Number of families with 1 girl = 814.
$P(E2) = \frac{\text{Number of families with 1 girl}}{\text{Total number of families}}$
$P(\text{1 girl}) = \frac{814}{1500}$
$P(\text{1 girl}) = \frac{\cancel{814}^{407}}{\cancel{1500}_{750}} = \frac{407}{750}$
$P(\text{1 girl}) \approx 0.5427$
The probability of a family having 1 girl is $\frac{407}{750}$.
(iii) Probability of a family having no girl.
Let E3 be the event that a family has no girl (0 girls).
Number of families with 0 girls = 211.
$P(E3) = \frac{\text{Number of families with 0 girls}}{\text{Total number of families}}$
$P(\text{No girl}) = \frac{211}{1500}$
$P(\text{No girl}) \approx 0.1407$
The probability of a family having no girl is $\frac{211}{1500}$.
Check whether the sum of these probabilities is 1.
Sum of probabilities = $P(\text{2 girls}) + P(\text{1 girl}) + P(\text{No girl})$
Sum of probabilities = $\frac{475}{1500} + \frac{814}{1500} + \frac{211}{1500}$
Sum of probabilities = $\frac{475 + 814 + 211}{1500}$
Numerator sum = $475 + 814 + 211 = 1289 + 211 = 1500$
$\begin{array}{cccc} & 4 & 7 & 5 \\ & 8 & 1 & 4 \\ + & 2 & 1 & 1 \\ \hline 1 & 5 & 0 & 0 \\ \hline \end{array}$Sum of probabilities = $\frac{1500}{1500} = 1$
The sum of these probabilities is indeed 1.
Question 3. Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.
Answer:
Given:
From Example 5, Chapter 14, a survey of 40 students about their birth months was conducted. The number of students born in each month is represented in a bar graph.
Total number of students surveyed = 40.
From the bar graph (or the underlying data for it):
Number of students born in January = 3
Number of students born in February = 4
Number of students born in March = 2
Number of students born in April = 2
Number of students born in May = 5
Number of students born in June = 1
Number of students born in July = 3
Number of students born in August = 7
Number of students born in September = 6
Number of students born in October = 4
Number of students born in November = 4
Number of students born in December = 3
Total students = $3+4+2+2+5+1+3+7+6+4+4+3 = 44$.
Wait, the example states there were 40 students. Let's re-check the counts from the example image.
Months (Jan-Dec): 3, 4, 2, 2, 5, 1, 3, 7, 6, 4, 4, 3. Sum = 3+4+2+2+5+1+3+7+6+4+4+3 = 44. There is a discrepancy between the stated total of 40 students and the sum of frequencies (44) in the bar graph from Example 5. I will proceed using the sum of frequencies from the bar graph as the actual total number of outcomes, as it represents the data collected. Let's assume the statement "40 students were asked" might have been intended to match the sum of frequencies. Based on the data in the bar graph: Total number of students = 44. Number of students born in August = 7.To Find:
The probability that a student of the class was born in August.
Solution:
Let E be the event that a student chosen at random was born in August.
Number of favourable outcomes for E (students born in August) = 7.
Total number of possible outcomes (total students) = 44 (based on the sum of frequencies in the bar graph).
The empirical probability of event E is:
$P(E) = \frac{\text{Number of students born in August}}{\text{Total number of students}}$
$P(\text{Born in August}) = \frac{7}{44}$
The probability that a student of the class was born in August is $\frac{7}{44}$.
Note: There is a discrepancy in the original problem statement of Example 5 in Chapter 14, where it mentions 40 students but the frequencies in the bar graph sum up to 44. We have used the sum of frequencies (44) as the total number of students for the probability calculation, as this represents the actual data presented in the graph.
Question 4. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:
| Outcome | 3 heads | 2 heads | 1 head | No head |
| Frequency | 23 | 72 | 77 | 28 |
If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Answer:
Given:
Three coins are tossed simultaneously 200 times. The frequencies of outcomes are:
| Outcome | Frequency |
| 3 heads | 23 |
| 2 heads | 72 |
| 1 head | 77 |
| No head | 28 |
| Total | 200 |
Total number of trials (tosses) = 200.
To Compute:
The probability of getting 2 heads when the three coins are tossed again.
Solution:
Let E be the event of getting 2 heads.
From the frequency distribution table, the number of times 2 heads came up is the frequency for the outcome "2 heads", which is 72.
Number of favourable outcomes for E = 72.
Total number of trials = 200.
The empirical probability of event E is:
$P(E) = \frac{\text{Number of times 2 heads came up}}{\text{Total number of trials}}$
$P(\text{2 heads}) = \frac{72}{200}$
$P(\text{2 heads}) = \frac{\cancel{72}^{36}}{\cancel{200}_{100}} = \frac{36}{100} = 0.36$
The probability of 2 heads coming up is 0.36 or $\frac{36}{100}$ or $\frac{9}{25}$.
Question 5. An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:
| Monthly income (in ₹) | Vehicles per family | |||
| 0 | 1 | 2 | Above 2 | |
| Less than 7000 | 10 | 160 | 25 | 0 |
| 7000 – 10000 | 0 | 305 | 27 | 2 |
| 10000 – 13000 | 1 | 535 | 29 | 1 |
| 13000 – 16000 | 2 | 469 | 59 | 25 |
| 16000 or more | 1 | 579 | 82 | 88 |
Suppose a family is chosen. Find the probability that the family chosen is
(i) earning ₹ 10000 – 13000 per month and owning exactly 2 vehicles.
(ii) earning ₹ 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not own any vehicle.
(iv) earning ₹ 13000 – 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Answer:
Given:
Data from a survey of 2400 families showing the relationship between income level and the number of vehicles owned.
| Monthly income (in $\textsf{₹}$) | Vehicles per family | ||||
| 0 | 1 | 2 | Above 2 | ||
| Less than 7000 | 10 | 160 | 25 | 0 | |
| 7000 – 10000 | 0 | 305 | 27 | 2 | |
| 10000 – 13000 | 1 | 535 | 29 | 1 | |
| 13000 – 16000 | 2 | 469 | 59 | 25 | |
| 16000 or more | 1 | 579 | 82 | 88 | |
| Row Totals | $10+160+25+0 = 195$ | $0+305+27+2 = 334$ | $1+535+29+1 = 566$ | $2+469+59+25 = 555$ | $1+579+82+88 = 750$ |
| Column Totals | $10+0+1+2+1 = 14$ | $160+305+535+469+579 = 2048$ | $25+27+29+59+82 = 222$ | $0+2+1+25+88 = 116$ | |
Let's recheck the sum of families: $10+160+25+0 = 195$ (Income < 7000)
$0+305+27+2 = 334$ (Income 7000-10000)
$1+535+29+1 = 566$ (Income 10000-13000)
$2+469+59+25 = 555$ (Income 13000-16000)
$1+579+82+88 = 750$ (Income $\geq$ 16000)
Sum of row totals = $195 + 334 + 566 + 555 + 750 = 2400$. This matches the total number of families.
Sum of column totals = $14 + 2048 + 222 + 116 = 2400$. This also matches.
Total number of families surveyed = 2400.
To Find:
The probabilities of the given events for a randomly chosen family.
Solution:
The empirical probability of an event E is $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$.
Total number of trials = 2400 (total number of families).
(i) Probability of earning $\textsf{₹}$ 10000 – 13000 per month and owning exactly 2 vehicles.
Let E1 be the event that a family earns $\textsf{₹}$ 10000 – 13000 and owns exactly 2 vehicles.
From the table, locate the row for "10000 – 13000" income and the column for "Vehicles per family = 2". The number of families in this category is 29.
Number of favourable outcomes for E1 = 29.
$P(E1) = \frac{\text{Number of families earning 10k-13k with 2 vehicles}}{\text{Total number of families}}$
$P(\text{10k-13k income and 2 vehicles}) = \frac{29}{2400}$
The probability is $\frac{29}{2400}$.
(ii) Probability of earning $\textsf{₹}$ 16000 or more per month and owning exactly 1 vehicle.
Let E2 be the event that a family earns $\textsf{₹}$ 16000 or more and owns exactly 1 vehicle.
From the table, locate the row for "16000 or more" income and the column for "Vehicles per family = 1". The number of families in this category is 579.
Number of favourable outcomes for E2 = 579.
$P(E2) = \frac{\text{Number of families earning $\geq$ 16k with 1 vehicle}}{\text{Total number of families}}$
$P(\text{$\geq$ 16k income and 1 vehicle}) = \frac{579}{2400}$
The probability is $\frac{579}{2400}$.
(iii) Probability of earning less than $\textsf{₹}$ 7000 per month and does not own any vehicle.
Let E3 be the event that a family earns less than $\textsf{₹}$ 7000 and owns 0 vehicles.
From the table, locate the row for "Less than 7000" income and the column for "Vehicles per family = 0". The number of families in this category is 10.
Number of favourable outcomes for E3 = 10.
$P(E3) = \frac{\text{Number of families earning < 7k with 0 vehicles}}{\text{Total number of families}}$
$P(\text{< 7k income and 0 vehicles}) = \frac{10}{2400} = \frac{1}{240}$
The probability is $\frac{1}{240}$.
(iv) Probability of earning $\textsf{₹}$ 13000 – 16000 per month and owning more than 2 vehicles.
Let E4 be the event that a family earns $\textsf{₹}$ 13000 – 16000 and owns more than 2 vehicles (i.e., Above 2 vehicles).
From the table, locate the row for "13000 – 16000" income and the column for "Vehicles per family = Above 2". The number of families in this category is 25.
Number of favourable outcomes for E4 = 25.
$P(E4) = \frac{\text{Number of families earning 13k-16k with > 2 vehicles}}{\text{Total number of families}}$
$P(\text{13k-16k income and > 2 vehicles}) = \frac{25}{2400}$
$P(\text{13k-16k income and > 2 vehicles}) = \frac{\cancel{25}^{1}}{\cancel{2400}_{96}} = \frac{1}{96}$
The probability is $\frac{1}{96}$.
(v) Probability of owning not more than 1 vehicle.
Let E5 be the event that a family owns not more than 1 vehicle (i.e., 0 vehicles or 1 vehicle).
To find the number of families owning not more than 1 vehicle, we sum the frequencies in the columns for "0" vehicles and "1" vehicle, across all income levels.
Number of families with 0 vehicles = 10 + 0 + 1 + 2 + 1 = 14.
Number of families with 1 vehicle = 160 + 305 + 535 + 469 + 579 = 2048.
Number of favourable outcomes for E5 = (Families with 0 vehicles) + (Families with 1 vehicle)
Number of favourable outcomes for E5 = $14 + 2048 = 2062$
$P(E5) = \frac{\text{Number of families with $\leq$ 1 vehicle}}{\text{Total number of families}}$
$P(\text{$\leq$ 1 vehicle}) = \frac{2062}{2400}$
$P(\text{$\leq$ 1 vehicle}) = \frac{\cancel{2062}^{1031}}{\cancel{2400}_{1200}} = \frac{1031}{1200}$
The probability is $\frac{1031}{1200}$.
Question 6. Refer to Table 14.7, Chapter 14.
(i) Find the probability that a student obtained less than 20% in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Answer:
Given:
The frequency distribution table of marks obtained by 90 students in a mathematics test (Table 14.7, Example 7, Chapter 14):
| Marks | Number of students (Frequency) |
| 0 - 20 | 7 |
| 20 - 30 | 10 |
| 30 - 40 | 10 |
| 40 - 50 | 20 |
| 50 - 60 | 20 |
| 60 - 70 | 15 |
| 70 - above (70-100) | 8 |
| Total | 90 |
Total number of students = 90.
To Find:
Probabilities based on the given data.
Solution:
The empirical probability of an event E is $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$.
Total number of trials = 90 (total number of students).
(i) Probability that a student obtained less than 20% marks.
Let E1 be the event that a student obtained less than 20% marks.
In the frequency distribution table, the marks are given as intervals. The interval "0 - 20" includes marks from 0 up to (but not including, assuming continuous intervals) 20. Since the lowest mark is 0, all students in this interval scored less than 20%.
Number of students who obtained less than 20% marks (in the 0-20 interval) = 7.
Number of favourable outcomes for E1 = 7.
$P(E1) = \frac{\text{Number of students with marks < 20}}{\text{Total number of students}}$
$P(\text{Marks < 20%}) = \frac{7}{90}$
The probability that a student obtained less than 20% marks is $\frac{7}{90}$.
(ii) Probability that a student obtained marks 60 or above.
Let E2 be the event that a student obtained marks 60 or above ($\geq 60$).
From the table, this includes students in the intervals where the lower limit is 60 or more.
These intervals are "60 - 70" and "70 - above (70-100)".
Number of students in the 60-70 interval = 15.
Number of students in the 70-above interval = 8.
Number of favourable outcomes for E2 = (Students in 60-70) + (Students in 70-above)
Number of favourable outcomes for E2 = $15 + 8 = 23$
$P(E2) = \frac{\text{Number of students with marks $\geq$ 60}}{\text{Total number of students}}$
$P(\text{Marks $\geq$ 60}) = \frac{23}{90}$
The probability that a student obtained marks 60 or above is $\frac{23}{90}$.
Question 7. To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table.
| Opinion | Number of students |
| like | 135 |
| dislike | 65 |
Find the probability that a student chosen at random
(i) likes statistics,
(ii) does not like it.
Answer:
Given:
Results of a survey of 200 students about their opinion on statistics.
| Opinion | Number of students (Frequency) |
| like | 135 |
| dislike | 65 |
| Total | 200 |
Total number of students surveyed = 200.
To Find:
The probability that a randomly chosen student likes statistics or does not like it.
Solution:
The empirical probability of an event E is $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$.
Total number of trials = 200 (total number of students).
(i) Probability that a student chosen at random likes statistics.
Let E1 be the event that a student likes statistics.
Number of students who like statistics = 135.
Number of favourable outcomes for E1 = 135.
$P(E1) = \frac{\text{Number of students who like statistics}}{\text{Total number of students}}$
$P(\text{Likes statistics}) = \frac{135}{200}$
$P(\text{Likes statistics}) = \frac{\cancel{135}^{27}}{\cancel{200}_{40}} = \frac{27}{40} = 0.675$
The probability that a student chosen at random likes statistics is $\frac{27}{40}$ or 0.675.
(ii) Probability that a student chosen at random does not like it.
Let E2 be the event that a student does not like statistics.
Number of students who dislike statistics = 65.
Number of favourable outcomes for E2 = 65.
$P(E2) = \frac{\text{Number of students who dislike statistics}}{\text{Total number of students}}$
$P(\text{Does not like statistics}) = \frac{65}{200}$
$P(\text{Does not like statistics}) = \frac{\cancel{65}^{13}}{\cancel{200}_{40}} = \frac{13}{40} = 0.325$
Alternatively, the event "does not like statistics" is the complement of the event "likes statistics".
$P(\text{Does not like statistics}) = 1 - P(\text{Likes statistics})$
$P(\text{Does not like statistics}) = 1 - \frac{135}{200} = \frac{200 - 135}{200} = \frac{65}{200} = \frac{13}{40}$
This confirms the result.
The probability that a student chosen at random does not like statistics is $\frac{13}{40}$ or 0.325.
Question 8. Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within $\frac{1}{2}$ km from her place of work?
Answer:
Given:
From Q.2, Exercise 14.2, the distance (in km) of 40 engineers from their residence to their place of work:
5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2, 7, 9, 7, 8, 3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12.
Total number of engineers (trials) = 40.
To Find:
The empirical probabilities for the given events.
Solution:
The empirical probability of an event E is $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$.
Total number of trials = 40.
(i) Probability that an engineer lives less than 7 km from her place of work.
Let E1 be the event that an engineer lives less than 7 km (< 7 km).
We count the number of engineers whose distance is less than 7 km from the given data:
5, 3, 3, 5, 3, 2, 2, 6, 6. Count = 9.
The values are 5, 3, 5, 3, 2, 2, 6, 6. Wait, let's recount carefully from the list.
5, 3, 10, 20, 25, 11, 13, 7, 12, 31 19, 10, 12, 17, 18, 11, 32, 17, 16, 2 7, 9, 7, 8, 3, 5, 12, 15, 18, 3 12, 14, 2, 9, 6, 15, 15, 7, 6, 12 Values < 7: 5, 3, 2, 3, 5, 3, 2, 6, 6. Count = 9. Number of favourable outcomes for E1 = 9.$P(E1) = \frac{\text{Number of engineers living < 7 km}}{\text{Total number of engineers}}$
$P(\text{Distance < 7 km}) = \frac{9}{40}$
$P(\text{Distance < 7 km}) = 0.225$
The probability is $\frac{9}{40}$ or 0.225.
(ii) Probability that an engineer lives more than or equal to 7 km from her place of work.
Let E2 be the event that an engineer lives more than or equal to 7 km ($\geq 7$ km).
We count the number of engineers whose distance is 7 km or more from the given data.
Alternatively, this event is the complement of the event in (i). The total number of engineers is 40, and 9 live less than 7 km away.
Number of engineers living $\geq 7$ km = Total engineers - Number of engineers living < 7 km
Number of engineers living $\geq 7$ km = $40 - 9 = 31$
Number of favourable outcomes for E2 = 31.
$P(E2) = \frac{\text{Number of engineers living $\geq$ 7 km}}{\text{Total number of engineers}}$
$P(\text{Distance $\geq$ 7 km}) = \frac{31}{40}$
$P(\text{Distance $\geq$ 7 km}) = 0.775$
The probability is $\frac{31}{40}$ or 0.775.
(iii) Probability that an engineer lives within $\frac{1}{2}$ km from her place of work.
Let E3 be the event that an engineer lives within $\frac{1}{2}$ km, which means the distance is less than or equal to 0.5 km ($ \leq 0.5$ km).
We check the given data to see if any engineer lives 0.5 km or less from work. The distances are given as whole numbers or 0, 1, 2... etc., from the context of the original problem values like 5, 3, 10... The smallest distance recorded is 2 km.
There are no engineers in the provided data whose distance is less than or equal to 0.5 km.
Number of favourable outcomes for E3 = 0.
$P(E3) = \frac{\text{Number of engineers living $\leq$ 0.5 km}}{\text{Total number of engineers}}$
$P(\text{Distance $\leq$ 0.5 km}) = \frac{0}{40} = 0$
The probability is 0.
Question 9. Activity: Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.
Answer:
Given:
This is an activity-based question. The data needs to be collected by the student.
To Find:
The probability that a randomly observed vehicle is a two-wheeler.
Solution:
Let's assume, for the purpose of demonstrating the solution structure, that a student carried out the activity and collected the following hypothetical data:
- Number of two-wheelers observed = 120
- Number of three-wheelers observed = 30
- Number of four-wheelers observed = 80
Total number of vehicles observed = Number of two-wheelers + Number of three-wheelers + Number of four-wheelers
Total number of vehicles observed = $120 + 30 + 80 = 230$
Total number of trials (vehicles observed) = 230.
Let E be the event that a randomly observed vehicle is a two-wheeler.
Number of favourable outcomes for E (number of two-wheelers) = 120.
Total number of trials = 230.
The empirical probability of event E is:
$P(E) = \frac{\text{Number of two-wheelers observed}}{\text{Total number of vehicles observed}}$
$P(\text{Two-wheeler}) = \frac{120}{230} = \frac{12}{23}$
Based on the hypothetical data, the probability that any one vehicle out of the total vehicles observed is a two-wheeler is $\frac{12}{23}$.
Instructions for the student performing the activity:
1. Choose a specific time interval (e.g., 15 minutes, 30 minutes) and a location (in front of the school gate).
2. Count and record the number of two-wheelers, three-wheelers, and four-wheelers that pass during this time interval.
3. Calculate the total number of vehicles observed by summing the counts for two-wheelers, three-wheelers, and four-wheelers.
4. Calculate the probability of a two-wheeler using the formula:
$P(\text{Two-wheeler}) = \frac{\text{Number of two-wheelers observed}}{\text{Total number of vehicles observed}}$
Question 10. Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.
Answer:
Given:
This is an activity-based question. The data needs to be collected from the students in your class.
To Find:
The probability that a randomly chosen student wrote a 3-digit number divisible by 3.
Solution:
Let's outline the steps to perform the activity and calculate the probability.
1. Ask every student in your class to write down any 3-digit number. A 3-digit number is an integer from 100 to 999, inclusive.
2. Collect all the written numbers from the students.
3. Count the total number of students who wrote a number. Let this be $N$. This is the total number of trials.
4. For each number written by a student, check if it is divisible by 3. A number is divisible by 3 if the sum of its digits is divisible by 3.
Example: If a student wrote 123, the sum of digits is $1+2+3=6$. Since 6 is divisible by 3, the number 123 is divisible by 3.
Example: If a student wrote 451, the sum of digits is $4+5+1=10$. Since 10 is not divisible by 3, the number 451 is not divisible by 3.
5. Count the number of students whose written number is divisible by 3. Let this be $M$. This is the number of favourable outcomes.
The empirical probability that a randomly chosen student wrote a number divisible by 3 is given by the formula:
$P(\text{Number divisible by 3}) = \frac{\text{Number of students who wrote a number divisible by 3}}{\text{Total number of students}}$
$P(\text{Number divisible by 3}) = \frac{M}{N}$
Example with hypothetical data:
Suppose there are $N = 30$ students in the class.
Suppose, after checking all the numbers, it is found that $M = 12$ students wrote a number that is divisible by 3.
In this hypothetical case, the probability would be:
$P(\text{Number divisible by 3}) = \frac{12}{30} = \frac{\cancel{12}^{2}}{\cancel{30}_{5}} = \frac{2}{5} = 0.4$
Remember to perform the activity in your class using the actual numbers written by your classmates to get the correct probability for your specific class.
Question 11. Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):
| 4.97 | 5.05 | 5.08 | 5.03 | 5.00 | 5.06 | 5.08 | 4.98 | 5.04 | 5.07 | 5.00 |
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Answer:
Given:
The actual weights (in kg) of flour in 11 bags:
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00.
Total number of bags = 11.
To Find:
The probability that a randomly chosen bag contains more than 5 kg of flour.
Solution:
Total number of trials (bags) = 11.
Let E be the event that a bag chosen at random contains more than 5 kg of flour.
We need to count the number of bags whose weight is strictly greater than 5 kg ($> 5$ kg) from the given data:
- 4.97 (Not > 5)
- 5.05 ( > 5)
- 5.08 ( > 5)
- 5.03 ( > 5)
- 5.00 (Not > 5)
- 5.06 ( > 5)
- 5.08 ( > 5)
- 4.98 (Not > 5)
- 5.04 ( > 5)
- 5.07 ( > 5)
- 5.00 (Not > 5)
The weights that are more than 5 kg are: 5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07.
Number of favourable outcomes for E = 7.
Total number of trials = 11.
The empirical probability of event E is:
$P(E) = \frac{\text{Number of bags containing > 5 kg}}{\text{Total number of bags}}$
$P(\text{Weight > 5 kg}) = \frac{7}{11}$
The probability that any of these bags chosen at random contains more than 5 kg of flour is $\frac{7}{11}$.
Question 12. In Q.5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 - 0.16 on any of these days.
Answer:
Given:
From Q.5, Exercise 14.2, the frequency distribution table for the concentration of sulphur dioxide in the air for 30 days:
| Concentration (in ppm) | Number of Days (Frequency) |
| 0.00 - 0.04 | 4 |
| 0.04 - 0.08 | 7 |
| 0.08 - 0.12 | 9 |
| 0.12 - 0.16 | 2 |
| 0.16 - 0.20 | 4 |
| 0.20 - 0.24 | 2 |
| Total | 30 |
Total number of days observed (trials) = 30.
To Find:
The probability of the concentration of sulphur dioxide being in the interval 0.12 - 0.16 on any of these days.
Solution:
Let E be the event that the concentration of sulphur dioxide is in the interval 0.12 - 0.16 ppm.
From the frequency distribution table, the number of days where the concentration was in the interval 0.12 - 0.16 is the frequency for this interval, which is 2.
Number of favourable outcomes for E = 2.
Total number of trials = 30.
The empirical probability of event E is:
$P(E) = \frac{\text{Number of days with concentration in 0.12 - 0.16}}{\text{Total number of days}}$
$P(\text{Concentration in 0.12 - 0.16}) = \frac{2}{30}$
$P(\text{Concentration in 0.12 - 0.16}) = \frac{\cancel{2}^{1}}{\cancel{30}_{15}} = \frac{1}{15}$
The probability of the concentration of sulphur dioxide being in the interval 0.12 - 0.16 on any of these days is $\frac{1}{15}$.
Question 13. In Q.1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Answer:
Given:
From Q.1, Exercise 14.2, the frequency distribution table for the blood groups of 30 students of Class VIII:
| Blood Group | Number of Students (Frequency) |
| A | 9 |
| B | 6 |
| O | 12 |
| AB | 3 |
| Total | 30 |
Total number of students = 30.
To Determine:
The probability that a randomly selected student has blood group AB.
Solution:
Let E be the event that a student chosen at random has blood group AB.
From the frequency distribution table, the number of students with blood group AB is 3.
Number of favourable outcomes for E = 3.
Total number of possible outcomes (total students) = 30.
The empirical probability of event E is:
$P(E) = \frac{\text{Number of students with blood group AB}}{\text{Total number of students}}$
$P(\text{Blood group AB}) = \frac{3}{30}$
$P(\text{Blood group AB}) = \frac{\cancel{3}^{1}}{\cancel{30}_{10}} = \frac{1}{10} = 0.1$
The probability that a student of this class, selected at random, has blood group AB is $\frac{1}{10}$ or 0.1.