NCERT solutions for class 9 Maths
Probability
Chapter 15
Exercise 15.1
Ex 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.
Solution:
Total number of trials = 30
∵ Number of times, the ball touched the boundary =6
∴ Number of times, the ball missed the boundary = 30 – 6 = 24
P (she did not hit a boundary) =24/30 =4/5 =0.8
Ex 15.1 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.
Solution:
Total number of families = 1500
(i) Number of families having 2 girls = 475
∴ P(2 girls) = 475/1500 = 19/60
Hence, the probability of families having two girls = 19 / 60.
(ii) ∵ Number of families having 1 girl = 814
∴ P (1 girl) = 814 / 1500 = 407 / 750
Hence, the probability of families having one girls = 407 / 750.
(iii) Number of families having no girl = 211
∴ P (no girl) = 211 / 1500
Hence, the probability of families having no girls = 211 / 1500.
Ex 15.1 Question 3.
In a particular section of class IX, 40 students were asked about the month of their birth and the following graph was prepared for the data so obtained.
Find the probability that a student of the class was born in August.
Solution:
Total number of students = 40
Number of students born in August = 6
∴ P (student born in august) = 6 / 40 = 3 / 20
Hence, the probability that the student born in august = 3 / 20.
Ex 15.1 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.
Solution:
Total number of tossed = 200
Number of getting 2 heads = 72
∴ P (2 heads) = 72 / 200 = 9 / 25
∴ Hence, the probability of getting two heads = 9 / 25
Ex 15.1 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.
Solution:
Total number of families = 2400
(i) ∵ Number of families earning ₹ 10000 – ₹ 13000 per month and owning exactly 2 vehicles = 29
∴ Probability = 29 / 2400
Hence, the probability of families earning ₹ 10000 – ₹ 13000 per month and owning exactly 2 vehicles = 29 / 2400
(ii) ∵ Number of families earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579
∴ Probability = 579 / 2400 = 193 / 800
Hence, the probability of families earning ₹ 16000 per month and owning exactly 1 vehicles = 193 / 800
(iii) ∵ Number of families earning less than Rs. 7000 per month and do not own any vehicle = 10
∴ Probability = 10 / 2400 = 1 / 240
Hence, the probability of families earning ₹ 7000 per month and does not own any vehicle = 1 / 240.
(iv) ∵ Number of families earning ₹ 13000 – ₹ 16000 per month and owning more than 2 vehicles = 25
∴ Probability = 25 / 2400 = 1 / 96
Hence, the probability of a family earning ₹ 13000 – ₹ 16000 per month and owning more than 2 vehicles = 1 / 96
(v) ∵ Number of families owning not more than 1 vehicle
= [Number of families having no vehicle] + [Number of families having only 1 vehicle]
= [10 + 1 + 2 + 1] + [160 + 305 + 535 + 469 + 579] = 14 + 2048 = 2062
∴ Probability = 2062 / 2400 = 1031 / 1200
Hence, the probability of a family owning not more than 1 vehicle =1031/1200.
Ex 15.1 Question 6.
A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows
0 – 20, 20 – 30, …, 60 – 70, 70 – 100. Then she formed the following table
| 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 |
(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.
Solution:
Total number of students = 90
(i) Number of students who obtained less than 20% marks = 7
Probability = 7 / 90
Hence, the probability of a student obtaining less than 20% marks = 7 / 90
(ii)Number of students who obtained marks 60 or above = [Number of students in class-interval 60 – 70] + [Number of students in the class interval 70 – above]
= 15 + 8 = 23
∴ Probability = 23 / 90
Hence,the probability of a student who obtained 23 marks 60 or above = 23/90
Ex 15.1 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.
Solution:
Total number of students = 200
(i) ∵ Number of students who like statistics = 135
∴ P (student likes statistics) = 135 / 200 = 27 / 40
Hence, the probability of selecting a student who likes statistics = 27 / 40
(ii) ∵ Number of students who do not like statistics = 65
∴ P (student who dislike statistics) = 60 / 200 = 13 / 40
Hence, the probability of selecting a student who dislikes statistics = 13 / 40.
Ex 15.1Question 8.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows
5 3 10 20 25 11 13 7 12 31 19 10 12 17 18 11 3 2
17 16 2 7 9 7 8 3 5 12 15 18 3 12 14 2 9 6
15 15 7 6 12
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 1/2 km from her place of work?
Solution:
Total number of engineers = 40
(i) ∵ Number of engineers who are living less than 7 km from their work place = 9
∴ P (an engineer who is living less than 7 km from her place of work )= 9 / 40
Hence, the probability of an engineer who is living less than 7 km from her place of work = 9 / 40
(ii) ∵ Number of engineers living at a distance more than or equal to 7 km from their work place = 31
∴ P (an engineer who is living at distance more than or equal to 7 km from her place of work)= 31 /40
Hence, the probability of an engineer who is living at distance more than or equal to 7 km from her place of work = 31 /40
(iii) ∵ The number of engineers living within 1 / 2 km from their work place = 0
∴ P (an engineer who is living within 1/2 km from her place of work) = 0/40= 0
Hence, the probability of an engineer who is living within 1/2 km from her place of work = 0
Ex 15.1 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?
Solution:
It is an activity. Students can do it themselves.
Ex 15.1 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 digit is divisible by 3.
Solution:
A class room activity for students.
Ex 15.1 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.
Solution:
Here, total number of bags = 11
∵ Number of bags having more than 5 kg of flour = 7
∴ Probability of a bag having more than 5 kg of flour = 7 / 11
Ex 15.1 Question 12.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows
0.03 0.08 0.08 0.09 0.04 0.17 0.16 0.05 0.02 0.06 0.18 0.20 0.11 0.08 0.12 0.13 0.22 0.07 0.08 0.01 0.10 0.06 0.09 0.18 0.11 0.07 0.05 0.07 0.01 0.04
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.
Solution:
Here, total number of days = 30
∵ The number of days on which the sulphur dioxide concentration is in the interval 0.12 – 0.16 = 2
∴ Probability of a day on which sulphur dioxide is in the interval 0.12 – 0.16 = 2 / 30 = 1 / 15.
Ex 15.1 Question 13.
The blood groups of 30 students of class VIII are recorded as follows
A, B, 0, 0, AB, 0, A, 0, B, A, 0, B, A, 0, 0, A, AB, 0, A, A, 0, 0, AB, B, A, B, 0
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.
Solution:
Here, total number of students = 30
∵ Number of students having blood group AB = 3
∴ Probability of a student whose blood group is AB = 3 / 30 = 1 / 10.