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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: $105 + 275 + 120 = 500$. 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 for which the record is available = 250.
Number of days the weather forecast was correct = 175.
(i) What is the probability that on a given day it was correct?
Solution
Let E1 be the event that the weather forecast was correct on a given day.
The total number of trials is the total number of days, which is 250.
The number of trials in which the event E1 happened (i.e., the forecast was correct) is 175.
The probability of an event is given by the formula:
$P(\text{Event}) = \frac{\text{Number of trials in which the event happened}}{\text{The total number of trials}}$
Therefore, the probability that the forecast was correct on a given day is:
$P(E_1) = \frac{\text{Number of days the forecast was correct}}{\text{Total number of days}}$
$P(E_1) = \frac{175}{250}$
To simplify the fraction, we can divide both the numerator and the denominator by their greatest common divisor, which is 25.
$P(E_1) = \frac{\cancel{175}^{7}}{\cancel{250}_{10}} = \frac{7}{10}$
$P(E_1) = 0.7$
So, the probability that the forecast was correct on a given day is 0.7.
(ii) What is the probability that it was not correct on a given day?
Solution
Let E2 be the event that the weather forecast was not correct on a given day.
First, we find the number of days the forecast was not correct.
Number of days the forecast was not correct = Total number of days - Number of days the forecast was correct
Number of days not correct = $250 - 175 = 75$
The number of trials in which the event E2 happened is 75.
The probability that the forecast was not correct on a given day is:
$P(E_2) = \frac{\text{Number of days the forecast was not correct}}{\text{Total number of days}}$
$P(E_2) = \frac{75}{250}$
Simplifying the fraction:
$P(E_2) = \frac{\cancel{75}^{3}}{\cancel{250}_{10}} = \frac{3}{10}$
$P(E_2) = 0.3$
So, the probability that the forecast was not correct on a given day is 0.3.
Alternate Solution for part (ii)
The event that the forecast was 'not correct' is the complement of the event that it was 'correct'.
The sum of the probabilities of an event and its complement is always 1.
$P(\text{correct}) + P(\text{not correct}) = 1$
From part (i), we found that $P(\text{correct}) = 0.7$.
Therefore,
$P(\text{not correct}) = 1 - P(\text{correct})$
$P(\text{not correct}) = 1 - 0.7 = 0.3$
This confirms our result.
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
Total number of tyres surveyed (total number of cases) = 1000.
The frequency distribution of the distance covered by the tyres is as follows:
| Distance (in km) | Frequency |
| less than 4000 | 20 |
| 4000 to 9000 | 210 |
| 9001 to 14000 | 325 |
| more than 14000 | 445 |
| Total | 1000 |
(i) What is the probability that it will need to be replaced before it has covered 4000 km?
Solution:
Let E1 be the event that a tyre needs to be replaced before covering 4000 km.
The total number of trials is the total number of tyres, which is 1000.
The number of favourable outcomes for E1 (tyres needing replacement before 4000 km) is 20.
The probability is calculated as:
$P(E_1) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
$P(E_1) = \frac{20}{1000} = \frac{2}{100} = \frac{1}{50}$
$P(E_1) = 0.02$
The probability that a tyre will need to be replaced before covering 4000 km is 0.02.
(ii) What is the probability that it will last more than 9000 km?
Solution:
Let E2 be the event that a tyre lasts more than 9000 km.
This means the tyre needs replacement either in the range '9001 to 14000 km' or 'more than 14000 km'.
The number of favourable outcomes for E2 is the sum of the frequencies for these two categories.
Number of favourable outcomes = (Frequency for '9001 to 14000') + (Frequency for 'more than 14000')
Number of favourable outcomes = $325 + 445 = 770$
The probability is:
$P(E_2) = \frac{770}{1000} = \frac{77}{100}$
$P(E_2) = 0.77$
The probability that a tyre will last more than 9000 km is 0.77.
(iii) What is the probability that it will need to be replaced after it has covered somewhere between 4000 km and 14000 km?
Solution:
Let E3 be the event that a tyre needs replacement between 4000 km and 14000 km.
This includes the categories '4000 to 9000 km' and '9001 to 14000 km'.
The number of favourable outcomes for E3 is the sum of the frequencies for these two categories.
Number of favourable outcomes = (Frequency for '4000 to 9000') + (Frequency for '9001 to 14000')
Number of favourable outcomes = $210 + 325 = 535$
The probability is:
$P(E_3) = \frac{535}{1000}$
$P(E_3) = 0.535$
The probability that a tyre will need replacement 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 are:
Test I: 69, Test II: 71, Test III: 73, Test IV: 68, Test V: 74
Total number of unit tests (total 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.
First, we count the number of tests where the student's marks were more than 70%.
- Test I: 69 (No)
- Test II: 71 (Yes)
- Test III: 73 (Yes)
- Test IV: 68 (No)
- Test V: 74 (Yes)
The student scored more than 70% in 3 tests.
Number of trials in which the event E happened = 3.
The probability of an event is calculated as:
$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
$P(E) = \frac{3}{5}$
$P(E) = 0.6$
Therefore, the probability that the student gets more than 70% marks in a unit test is $\frac{3}{5}$ or 0.6.
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 the number of accidents in one year.
Total number of drivers (total trials) = 2000.
To Find
The probabilities of the specified events for a randomly chosen driver.
Solution
The probability of an event is given by the formula:
$P(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
(i) Probability of being 18-29 years of age and having exactly 3 accidents in one year.
From the table, the number of drivers in the age group 18-29 who had exactly 3 accidents is 61.
Number of favourable outcomes = 61.
$P(\text{18-29 years and 3 accidents}) = \frac{61}{2000} = 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.
We need to find the total number of drivers in the 30-50 age group who had 1, 2, 3, or over 3 accidents.
Number of favourable outcomes = $125 + 60 + 22 + 18 = 225$.
$P(\text{30-50 years and $\geq$ 1 accident}) = \frac{225}{2000} = \frac{9}{80} = 0.1125$
The probability is 0.1125.
(iii) Probability of having no accidents in one year.
We need to find the total number of drivers across all age groups who had 0 accidents.
Number of favourable outcomes = $440 + 505 + 360 = 1305$.
$P(\text{No accidents}) = \frac{1305}{2000} = \frac{261}{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 of weights of 38 students (from Table 14.3, 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 (total trials) = 38.
(i) Find the probability that the weight of a student in the class lies in the interval 46-50 kg.
Let E be the event that a student's weight is in the interval 46-50 kg.
From the table, the number of students in the 46-50 kg interval is 3.
Number of favourable outcomes = 3.
$P(E) = \frac{\text{Number of students in 46-50 kg interval}}{\text{Total number of students}} = \frac{3}{38}$
The probability is $\frac{3}{38}$.
(ii) Give two events in this context, one having probability 0 and the other having probability 1.
Event with probability 0 (Impossible Event):
An event that cannot happen based on the given data.
Example: The event that a student weighs less than 31 kg. According to the table, the lowest weight category starts at 31 kg, so no student has a weight less than 31 kg.
The probability is $\frac{0}{38} = 0$.
Event with probability 1 (Sure Event):
An event that is certain to happen.
Example: The event that a student weighs at most 75 kg. Since the highest weight class is 71-75 kg, all 38 students have weights within this overall range (31 to 75 kg).
The probability is $\frac{38}{38} = 1$.
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
The number of seeds that germinated from a sample of 50 seeds, taken from 5 different bags.
Bag 1: 40, Bag 2: 48, Bag 3: 42, Bag 4: 39, Bag 5: 41.
Total number of bags (total trials) = 5.
To Find
The probabilities of the specified events.
Solution
(i) Probability of germination of more than 40 seeds in a bag.
We look for bags where the number of germinated seeds is > 40. These are Bag 2 (48), Bag 3 (42), and Bag 5 (41).
Number of favourable outcomes = 3.
$P(\text{more than 40 seeds}) = \frac{3}{5} = 0.6$
The probability is 0.6.
(ii) Probability of germination of 49 seeds in a bag.
We look for bags where exactly 49 seeds germinated. None of the bags had 49 germinated seeds.
Number of favourable outcomes = 0.
$P(\text{49 seeds}) = \frac{0}{5} = 0$
The probability is 0.
(iii) Probability of germination of more than 35 seeds in a bag.
We look for bags where the number of germinated seeds is > 35. These are Bag 1 (40), Bag 2 (48), Bag 3 (42), Bag 4 (39), and Bag 5 (41).
All 5 bags satisfy this condition.
Number of favourable outcomes = 5.
$P(\text{more than 35 seeds}) = \frac{5}{5} = 1$
The probability 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 = 30.
Number of times a boundary was hit = 6.
To Find
The probability that she did not hit a boundary.
Solution
First, we find the number of balls in which she did not hit a boundary.
Number of balls without a boundary = Total balls played - Number of boundaries hit
Number of balls without a boundary = $30 - 6 = 24$.
Let E be the event that the batswoman did not hit a boundary.
The total number of trials is 30.
The number of outcomes favourable to event E is 24.
The probability of event E is given by:
$P(E) = \frac{\text{Number of outcomes favourable to E}}{\text{Total number of trials}}$
$P(\text{did not hit a boundary}) = \frac{24}{30} = \frac{4}{5}$
$P(\text{did not hit a boundary}) = 0.8$
Therefore, the probability that she did not hit a boundary is $\frac{4}{5}$ or 0.8.
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
A survey of 1500 families with 2 children.
Total number of families = 1500.
Number of families with 2 girls = 475.
Number of families with 1 girl = 814.
Number of families with no girl = 211.
To Compute
The probabilities for each case and to check if their sum is 1.
Solution
(i) Probability of a family having 2 girls
$P(\text{2 girls}) = \frac{\text{Number of families with 2 girls}}{\text{Total number of families}} = \frac{475}{1500} = \frac{19}{60}$
(ii) Probability of a family having 1 girl
$P(\text{1 girl}) = \frac{\text{Number of families with 1 girl}}{\text{Total number of families}} = \frac{814}{1500} = \frac{407}{750}$
(iii) Probability of a family having no girl
$P(\text{no girl}) = \frac{\text{Number of families with no girl}}{\text{Total number of families}} = \frac{211}{1500}$
Checking the sum of probabilities:
Sum = $P(\text{2 girls}) + P(\text{1 girl}) + P(\text{no girl})$
Sum = $\frac{475}{1500} + \frac{814}{1500} + \frac{211}{1500}$
Sum = $\frac{475 + 814 + 211}{1500} = \frac{1500}{1500} = 1$
Yes, the sum of the probabilities is 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
The problem refers to the data from Example 5, Section 14.4, Chapter 14, which is represented by the provided bar graph (Fig. 14.1).
From the bar graph:
- The total number of students is the sum of the students born in each month: $3 (\text{Jan}) + 4 (\text{Feb}) + 2 (\text{Mar}) \ $$ + 2 (\text{Apr}) \ $$ + 5 (\text{May}) \ $$ + 1 (\text{Jun}) \ $$ + 2 (\text{Jul}) + 6 (\text{Aug}) + 3 (\text{Sep}) \ $$ + 4 (\text{Oct}) + 4 (\text{Nov}) \ $$ + 4 (\text{Dec}) = 40$.
- The total number of students (total outcomes) = 40.
- The number of students born in the month of August = 6.
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.
The total number of trials is the total number of students, which is 40.
The number of favourable outcomes is the number of students born in August, which is 6.
The probability of an event E is given by the formula:
$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
Substituting the values into the formula:
$P(\text{student born in August}) = \frac{\text{Number of students born in August}}{\text{Total number of students}}$
$P(E) = \frac{6}{40}$
Simplifying the fraction:
$P(E) = \frac{\cancel{6}^{3}}{\cancel{40}_{20}} = \frac{3}{20}$
As a decimal, this is:
$P(E) = 0.15$
Therefore, the probability that a student of the class was born in August is $\frac{3}{20}$ or 0.15.
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
Total number of times three coins were tossed = 200.
Frequency of getting 2 heads = 72.
To Compute
The probability of 2 heads coming up in a single toss.
Solution
Let E be the event of getting 2 heads.
The total number of trials is 200.
The number of times the event E (getting 2 heads) occurred is 72.
The empirical probability of this event is:
$P(E) = \frac{\text{Frequency of 2 heads}}{\text{Total number of tosses}}$
$P(E) = \frac{72}{200} = \frac{36}{100} = \frac{9}{25}$
$P(E) = 0.36$
The probability of 2 heads coming up is $\frac{9}{25}$ or 0.36.
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
A survey of 2400 families linking income to the number of vehicles.
Total number of families = 2400.
To Find
The probabilities of the specified events.
Solution
The total number of outcomes is 2400.
(i) Earning $\textsf{₹}$ 10000 – 13000 per month and owning exactly 2 vehicles.
From the table, the number of such families is 29.
$P(\text{Event i}) = \frac{29}{2400}$
(ii) Earning $\textsf{₹}$ 16000 or more per month and owning exactly 1 vehicle.
From the table, the number of such families is 579.
$P(\text{Event ii}) = \frac{579}{2400} = \frac{193}{800}$
(iii) Earning less than $\textsf{₹}$ 7000 per month and does not own any vehicle.
From the table, the number of such families is 10.
$P(\text{Event iii}) = \frac{10}{2400} = \frac{1}{240}$
(iv) Earning $\textsf{₹}$ 13000 – 16000 per month and owning more than 2 vehicles.
From the table, the number of such families is 25.
$P(\text{Event iv}) = \frac{25}{2400} = \frac{1}{96}$
(v) Owning not more than 1 vehicle.
This means owning 0 vehicles or 1 vehicle. We sum the total number of families in these two categories.
Number of families with 0 vehicles = $10 + 0 + 1 + 2 + 1 = 14$
Number of families with 1 vehicle = $160 + 305 + 535 + 469 + 579 = 2048$
Total families owning not more than 1 vehicle = $14 + 2048 = 2062$
$P(\text{Event v}) = \frac{2062}{2400} = \frac{1031}{1200}$
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
Data from Table 14.7, Chapter 14, regarding the marks of 90 students.
| Marks | Number of students |
| 0 - 20 | 7 |
| 20 - 30 | 10 |
| 30 - 40 | 10 |
| 40 - 50 | 20 |
| 50 - 60 | 20 |
| 60 - 70 | 15 |
| 70 - above | 8 |
| Total | 90 |
Total number of students = 90.
To Find
The specified probabilities.
Solution
Total number of outcomes = 90.
(i) Probability that a student obtained less than 20% in the mathematics test.
The number of students who scored less than 20% corresponds to the "0 - 20" interval.
Number of favourable outcomes = 7.
$P(\text{marks < 20%}) = \frac{7}{90}$
The probability is $\frac{7}{90}$.
(ii) Probability that a student obtained marks 60 or above.
This includes students in the "60 - 70" and "70 - above" intervals.
Number of favourable outcomes = $15 + 8 = 23$.
$P(\text{marks $\geq$ 60}) = \frac{23}{90}$
The probability 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
A survey of 200 students about their opinion on the subject statistics.
Number of students who like statistics = 135.
Number of students who dislike statistics = 65.
Total number of students = 200.
To Find
The specified probabilities.
Solution
Total number of outcomes = 200.
(i) Probability that a student likes statistics.
Number of favourable outcomes = 135.
$P(\text{likes statistics}) = \frac{135}{200} = \frac{27}{40}$
The probability is $\frac{27}{40}$.
(ii) Probability that a student does not like it.
Number of favourable outcomes = 65.
$P(\text{does not like it}) = \frac{65}{200} = \frac{13}{40}$
The probability is $\frac{13}{40}$.
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
The data from Q.2, Exercise 14.2, which lists the distance (in km) from residence to work for 40 engineers:
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 = 40.
To Find
The specified empirical probabilities.
Solution
Total number of outcomes = 40.
(i) Probability that an engineer lives less than 7 km from her place of work.
We count the number of engineers whose distance is < 7 km: 5, 3, 2, 3, 5, 3, 2, 6, 6. There are 9 such engineers.
Number of favourable outcomes = 9.
$P(\text{distance < 7 km}) = \frac{9}{40}$
The probability is $\frac{9}{40}$.
(ii) Probability that an engineer lives more than or equal to 7 km from her place of work.
This event is the complement of the event in part (i).
Number of favourable outcomes = Total engineers - Number of engineers living < 7 km = $40 - 9 = 31$.
$P(\text{distance $\geq$ 7 km}) = \frac{31}{40}$
The probability is $\frac{31}{40}$.
(iii) Probability that an engineer lives within $\frac{1}{2}$ km from her place of work.
This means the distance is $\leq 0.5$ km. Looking at the data, the smallest distance is 2 km. There are no engineers living within 0.5 km of their workplace.
Number of favourable outcomes = 0.
$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:
Activity-Based Question
This question requires the student to perform an activity and collect data. The answer will depend on the data collected.
Methodology
1. Observation: Stand in front of your school gate for a fixed period (e.g., 30 minutes).
2. Data Collection: Use a tally mark sheet to count the number of two-wheelers, three-wheelers, and four-wheelers that pass by.
3. Calculation:
- Let the number of two-wheelers be $N_{2W}$.
- Let the number of three-wheelers be $N_{3W}$.
- Let the number of four-wheelers be $N_{4W}$.
Total number of vehicles observed (Total trials), $N_{Total} = N_{2W} + N_{3W} + N_{4W}$.
The probability of observing a two-wheeler is:
$P(\text{two-wheeler}) = \frac{\text{Number of two-wheelers}}{\text{Total number of vehicles}} = \frac{N_{2W}}{N_{Total}}$
Example (Hypothetical Data)
Suppose during a 30-minute interval, you observe:
- Two-wheelers ($N_{2W}$) = 75
- Three-wheelers ($N_{3W}$) = 20
- Four-wheelers ($N_{4W}$) = 45
Total vehicles ($N_{Total}$) = $75 + 20 + 45 = 140$.
The probability of observing a two-wheeler would be:
$P(\text{two-wheeler}) = \frac{75}{140} = \frac{15}{28}$
In this hypothetical case, the probability is $\frac{15}{28}$.
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:
Activity-Based Question
This is an activity where the result depends on the data collected from the students in a specific class.
Methodology
1. Data Collection: Ask every student in your class to write a 3-digit number (from 100 to 999).
2. Count Total Students: Count the total number of students who participated. This is the total number of trials. Let's call this $N$.
3. Check for Divisibility by 3: For each number, add its digits. If the sum of the digits is divisible by 3, then the original number is divisible by 3.
4. Count Favourable Outcomes: Count how many of the numbers are divisible by 3. Let's call this count $M$.
5. Calculate Probability:
$P(\text{number is divisible by 3}) = \frac{\text{Number of students with a number divisible by 3}}{\text{Total number of students}} = \frac{M}{N}$
Theoretical Probability (For Comparison)
The total number of 3-digit numbers is $999 - 100 + 1 = 900$.
The 3-digit numbers divisible by 3 are 102, 105, ..., 999. This is an arithmetic progression.
Let $a = 102$, $d = 3$, and $a_n = 999$.
$a_n = a + (n-1)d$
$999 = 102 + (n-1)3$
$897 = (n-1)3$
$299 = n-1 \implies n = 300$.
There are 300 three-digit numbers divisible by 3.
The theoretical probability is $\frac{300}{900} = \frac{1}{3}$.
Your empirical probability from the class activity should be close to this theoretical value of $\frac{1}{3}$.
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 are:
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
Let E be the event that a randomly chosen bag contains more than 5 kg of flour.
Total number of trials = 11.
We need to count the number of bags with weight > 5.00 kg.
The weights greater than 5.00 kg are: 5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07.
The number of such bags is 7.
Number of favourable outcomes = 7.
$P(E) = \frac{\text{Number of bags with more than 5 kg}}{\text{Total number of bags}}$
$P(E) = \frac{7}{11}$
The probability that a randomly chosen bag 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
The frequency distribution table for the concentration of sulphur dioxide for 30 days (from Q.5, Exercise 14.2).
| Concentration of SO2 (in ppm) | Number of Days |
| 0.00 - 0.04 | 4 |
| 0.04 - 0.08 | 9 |
| 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 = 30.
To Find
The probability that the concentration of sulphur dioxide is in the interval 0.12 - 0.16.
Solution
Let E be the event that the concentration is in the interval 0.12 - 0.16.
Total number of trials = 30.
From the table, the number of days the concentration was in the interval 0.12 - 0.16 is 2.
Number of favourable outcomes = 2.
$P(E) = \frac{\text{Number of days in interval 0.12 - 0.16}}{\text{Total number of days}}$
$P(E) = \frac{2}{30} = \frac{1}{15}$
The probability 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
The frequency distribution table for the blood groups of 30 students (from Q.1, Exercise 14.2).
| Blood Group | Number of Students |
| A | 9 |
| B | 6 |
| AB | 3 |
| O | 12 |
| 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 has blood group AB.
Total number of trials = 30.
From the table, the number of students with blood group AB is 3.
Number of favourable outcomes = 3.
$P(E) = \frac{\text{Number of students with blood group AB}}{\text{Total number of students}}$
$P(E) = \frac{3}{30} = \frac{1}{10}$
$P(E) = 0.1$
The probability that a randomly selected student has blood group AB is $\frac{1}{10}$ or 0.1.