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NCERT Solutions for Class 9 Maths Chapter 15 Probability

NCERT Solutions for Class 9 Maths Chapter 15 Probability Statistics are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 15 Probability.

Board CBSE
Textbook NCERT
Class Class 9
Subject Maths
Chapter Chapter 15
Chapter Name  Probability
Exercise  Ex 15.1
Number of Questions Solved 13
Category NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 15 Probability

Ex 15.1 Class 9 Maths Question 1.
In a cricket match, a bats woman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Solution:
Given, a bats woman play 30 balls, therefore total number of trials = 30 and number of events of hitting the boundary = 6
Now, number of balls in which she is not hitting the boundary
= 30 – 6 = 24
∴ The probability that she did not hit a boundary = \cfrac { 24 }{ 30 } = \cfrac { 4 }{ 5 }

Ex 15.1 Class 9 Maths 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:
Let E0, E1 and E2 be the event of getting no girl, 1 girl and 2 girls.
NCERT Solutions for Class 9 Maths Chapter 15 Probability 1
NCERT Solutions for Class 9 Maths Chapter 15 Probability 2

Ex 15.1 Class 9 Maths Question 3.
In a particular section of class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained.
NCERT Solutions for Class 9 Maths Chapter 15 Probability 3
Find the probability that a student of the class was born in August.
Solution:
Total number of students in class IX, n(S) = 40
Number of students born in the month of August, n(E) = 6
Probability, that the students of the class was born in August
=  \cfrac { n(E) }{ n(S) } = \cfrac { 6 }{ 40 } = \cfrac { 3 }{ 20 }

Ex 15.1 Class 9 Maths 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:
In tossing of three coins, getting two heads comes out 72 times. i.e.,   n(E) = 72
The total number of tossed three coins n(s) = 200
∴ Probability of 2 heads coming up =  \cfrac { n(E) }{ n(S) } = \cfrac { 72 }{ 200 } = \cfrac { 9 }{ 25 }

Ex 15.1 Class 9 Maths 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 :
Solution:

Montly income(in Rs)
 Vehicles per Family
Less Than 7000 0
10		 1
160		 2
25		 Above 2
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 Rs. 10000 -13000 per month and owning exactly 2 vehicles.
(ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than Rs. 7000 per month and does not own any vehicle.
(iv) earning Rs. 13000 -16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution:
Total number of families selected by the organisation, n(S) = 2400
(i) The number of families earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles,
n(E1) = 29
NCERT Solutions for Class 9 Maths Chapter 15 Probability 4

Ex 15.1 Class 9 Maths Question 6.
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:
(i) Total number of students in a class, n(S) = 90
The number of students less than 20% lies in the interval 0 – 20,
i.e., n(E) = 7
∴ The probability, that a student obtained less than 20% in the Mathematics
test = \cfrac { n(E) }{ n(S) } = \cfrac { 23 }{ 90 }

(ii) The number of students obtained marks 60 or above lies in the marks interval 60 – 70 and 70-above
i.e., n(F) =15+8 = 23
∴  The probability that a student obtained marks 60 or above = \cfrac { n(E) }{ n(S) } = \cfrac { 23 }{ 90}

Ex 15.1 Class 9 Maths 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, n(S) = 200
(i) The number of students who like Statistics n(E) = 135
∴ The probability, that the student like Statistics
\cfrac { n(F) }{ n(S) } = \cfrac { 135 }{ 200} = \cfrac { 27 }{ 40}

(ii) The number of student Who does not like Statistics, n (E) = 65
∴ The probability, that the student does not like Statistics = \cfrac { n(F) }{ n(S) } = \cfrac { 65 }{ 200} = \cfrac { 13 }{ 40}

Ex 15.1 Class 9 Maths Question 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 32 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 \cfrac { 1 }{ 2 } km from her place of work?
Solution:
Total number of engineers lives, n(S) = 40
(i) The number of engineers whose residence is less than 7 km from their place, n(E) = 9
∴ The probability that an engineer lives less than 7 km from their place of work
= \cfrac { n(E) }{ n(S) } =  \cfrac { 9 }{ 40}

(ii) The number of engineers whose residence is more than or equal to 7 km from their place of work, n(F) = 40 – 9 = 31
∴ The probability, that an engineer lives more than or equal to 7 km from their place of work
= \cfrac { n(F) }{ n(S) } =  \cfrac { 31 }{ 40}

(iii) The number of engineers whose residence within \cfrac { 1 }{ 2} km from their place of work, i.e., n(G) = 0
∴ The probability, that an engineer lives within \cfrac { 1 }{ 2} km from their place
= \cfrac { n(G) }{ n(S) } =  \cfrac { 0 }{ 40} =0

Ex 15.1 Class 9 Maths 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:
After observing in front of the school gate in time interval 6 : 30 to 7 : 30 am respective frequencies of different types of vehicles are :

Types of Vehicles Frequency
Two wheelers 550
Three wheelers 250
Four wheelers 80

∴  Total number of vehicle, n(S) = 550 + 250 + 80 = 880
Number of two-wheelers, n(E) = 550
robability of observing two-wheelers = \cfrac { n(E) }{ n(S) } =  \cfrac { 550 }{ 880} =  \cfrac { 5 }{ 8}

Ex 15.1 Class 9 Maths 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.
Solution:
Suppose, there are 40 students in a class.
∴ The probability of selecting any of the student = \cfrac { 40 }{ 40 } = 1
A three digit number start from 100 to 999
Total number of three digit numbers = 999 – 99 = 900
∴ Multiple of 3 in three digit numbers = (102, 105,…, 999)
∴ Number of multiples of 3 in three digit number = \cfrac { 900 }{ 3 }  = 300
i.e.,  n(E) = 300
∴  The probability that the number written by her/him is divisible by 3
\cfrac { n(E) }{ n(S) } =  \cfrac { 300 }{ 900 } = \cfrac { 1 }{ 3 }

Ex 15.1 Class 9 Maths 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:
The total number of wheat flour bags, n(S) = 11
Bags, which contains more than 5 kg of flour, (E)
= {5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07,}
∴ n(E)=7
∴  Required probability = \cfrac { n(E) }{ n(S) } =  \cfrac { 7 }{ 11 }

Ex 15.1 Class 9 Maths 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:
Now, we prepare a frequency distribution table

Interval Frequency
0.01-0.04 5
0.04-0.08 11
0.08-0.12 7
0.12-0.16 2
0.16-0.20 4
0.20-0.24 1
Total 30

The total number of days for data, to prepare sulphur dioxide, n(S) = 30
The frequency of the sulphur dioxide in the interval 0.12-0.16, n(E) = 2
∴ Required probability = \cfrac { n(E) }{ n(S) } =  \cfrac { 2 }{ 30 } = \cfrac { 1 }{ 15 }

Ex 15.1 Class 9 Maths Question 13.
The blood groups of 30 students of class VIII are recorded as follows:
A,B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB,
B,A, O, B, A, B, O
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 randon, has blood group AB.

Blood group Frequency
A 9
B 6
O 12
AB 3
Total 30

The total number of students in class VIII, n(S) = 30
The number of students who have blood group AB, n(E) = 3
∴ The probability that a student has a blood group
AB = \cfrac { n(E) }{ n(S) } =  \cfrac { 3 }{ 30 } \cfrac { 1 }{ 10 }

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