These NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 15 Probability 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.

Solution:

Total number of balls she plays = 30

Total number of balls that she hits a boundary = 6

The total number of balls that she does not hit a boundary = 30 – 6 = 24 balls.

Let us denote the event that she does not hit a ball is E: So,

P(E) = \(\frac{24}{30}=\frac{4}{5}\)

Question 2.

1500 families with two children were selected randomly, and the following data were recorded.

Compute the probability of a family, chosen at random having.

(i) 2 girls

(ii) 1 girl

(iii) No girls

Also, check whether the sum of these probabilities is 1.

Solution:

Total number of families = 1500

(i) Total number of families which have 2 girls are 475.

∴ P(2 girls in a family) = \(\frac{475}{1500}=\frac{19}{60}\)

(ii) Total number of families which have 1 girl is 814.

∴ P(1 girl in a family) = \(\frac{814}{1500}=\frac{407}{750}\)

(iii) Total number of families which ahve no girl is 211.

∴ P(no girl in a family) = \(\frac{211}{1500}\)

Check:

Sum of probabilities = \(\frac{9}{60}+\frac{407}{750}+\frac{211}{1500}\)

= \(\frac{475+814+211}{1500}\)

= \(\frac{1500}{1500}\)

= 1

Question 3.

Refer to example 5, section 14.4, chapter 14 final the probability that a student of the class was born in August.

Solution:

Total number of students who were born in August = 6

Total number of students = 40

Therefore,

P(Number of students bom in August) = \(\frac{6}{40}\) = \(\frac{3}{20}\)

Question 4.

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.

Solution:

Since the coin is tossed 200 times, so the total number of trials is 200.

Let us call the events of getting 2 heads is E.

Then the number of times E happens, i.e. two heads coming up is 72.

So, P(E) = \(\frac{72}{200}=\frac{9}{25}\)

Question 5.

An organization selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below:

Suppose a family is chosen. Find the probability that the family chosen is:

(i) earning Rs. 10,000 – 13,000 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. 13,000 – 16,000 per month and owning more than 2 vehicles.

(v) owning not more than 1 vehicle.

Solution:

The total number of families = 2400.

(i) The number of families earning Rs. 10,000 – 13,000 per month and owning exactly two vehicles is 29.

So, P(family earning Rs. 10,000 – 13,000 per month and owning exactly two vehicles) = \(\frac{29}{2400}\)

(ii) The number of families earning Rs. 16,000 or more per month and owning exactly one vehicle is 579.

So, P(earning Rs. 16,000 or more per month and owning exactly one vehicle) = \(\frac{579}{2400}\)

(iii) The number of families earning less than Rs. 7000 per month and does not own any vehicle is 10.

So, P(earning less than Rs. 7000 per month and does not own any vehicle) = \(\frac{10}{2400}=\frac{1}{240}\)

(iv) The number of families earning Rs. 13,000 – 16,000 per month and owning more than two vehicles is 25.

So, P(earning Rs. 13,000 – 16,000 per month and owning more than two vehicles) = \(\frac{25}{2400}=\frac{1}{96}\)

(v) The number of families owning not more than 1 vehicle is 2062 (family owning exactly one vehicle + family owning no vehicle)

So, P(owning not more than one vehicle) = \(\frac{2062}{2400}=\frac{1031}{1200}\)

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.

Solution:

(i) Total number of students = 90

Number of students who gets less than 20% marks in mathematics test = 7.

∴ P(student obtained less than 20% marks) = \(\frac{7}{90}\)

(ii) Number of student getting 60 or above marks is 23.

∴ P(student obtained 60 or above marks) = \(\frac{23}{90}\)

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.

Find the probability that a student is chosen at random

(i) like statistics

(ii) does not like it.

Solution:

The total number of students = 200.

(i) Total number of students like statistics = 135

So, P(like statistics) = \(\frac{135}{200}=\frac{27}{40}\)

(ii) Total number of students does not like statistics = 65.

So, P(does not like statistics) = \(\frac{65}{200}=\frac{13}{40}\)

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?

Solution:

(i) Total number of female engineers whose residence to their place of work given is 40.

The number of engineers whose residence is less than 7 km from their place of work = 9.

∴ P(less than 7 km from her place of work) = \(\frac{9}{40}\)

(ii) Number of engineer who’s residence is more than or equal to 7 km from her place of work = 31.

∴ P(more than or equal to 7 km from place of work) = \(\frac{31}{40}\)

(iii) Number of engineer who’s residence is within \(\frac {1}{2}\) km. from her place of work = 0.

∴ P(within \(\frac {1}{2}\) km. from place of work) = \(\frac {0}{40}\) = 0

Question 9.

Activity: Note the frequency of two-wheelers, three-wheelers, and four-wheelers going past during the 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:

Let the person saw the vehicles for half an hour.

The number of vehicles seen by the person.

Two wheelers = 5

Three wheelers = 10

Four wheelers = 15

So,total number of space = 5 + 10 + 15 = 30

Number of events when two wheeler is seen = 5

Probability = \(\frac{\text { no. of event }}{\text { no. of space }}\)

= \(\frac{5}{30}\)

= \(\frac{1}{5}\)

Question 10.

Activity: Ask all the students in your class to write a three-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:

The total number of students in my class = 40.

Out of 40 students, 10 students write a three-digit number which is divisible by 3.

∴ P(student write a number which is divisible by 3) = \(\frac{10}{40}=\frac{1}{4}\)

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 is 11.

The total number of wheat flour bags which contain more than 5 kg. of flour is 7.

∴ P(bags contain more than 5 kg of flour) = \(\frac{7}{11}\)

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.

Solution:

The total number of observations = 30.

Number of observation lie in the interval 0.11 – 0.16 is 2.

∴ P(The concentration of sulphur dioxide in the interval 0.12 – 0.16 = \(\frac{2}{30}=\frac{1}{15}\)

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.

Solution:

Total number of students whose blood group are recorded = 30.

Number of students which have blood group AB is 3

∴ P(student has blood group AB) = \(\frac{3}{30}=\frac{1}{10}\)