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Class 9 NCERT Solutions- Chapter 15 Probability – Exercise 15.1
• Last Updated : 25 Jan, 2021

### 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:

Data given in the question:
Total number of balls batswoman plays = 30
Numbers of boundary hit by batswoman = 6
To find number of time batswoman didn’t hit boundary , we will substract
⇒(Total number of balls batswoman plays) – (Numbers of boundary hit by batswoman)
⇒ 30 – 6 = 24
Probability of that she did’t hit a boundary = 24/30 = 4/5

### (i) 2 girls (ii) 1 girl (iii) No girl

Solution:

According to question
Total numbers of families given in the question 1500
(i) Numbers of families having 2 girls = 475
Probability of choosen 2 girls = Numbers of families having 2 girls / Total numbers of families
= 475/1500 = 19/60
Probability of choosen 2 girls is 19/60

(ii) Numbers of families having 1 girls = 814
Probability of choosen 1 girl = Numbers of families having 1 girl / Total numbers of families
= 814/1500 = 407/750
Probability of choosen 1 girl is 407/750

(iii) Numbers of families having 2 girls = 211
Probability of choosen 0 girl = Numbers of families having 0 girls/Total numbers of families
= 211/1500

Sum of the probability = (19/60)+(407/750)+(211/1500)
= (475+814+211)/1500
= 1500/1500 = 1

Yes, the sum of these 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.

Solution:

According to questions:
Total number of students in the class in the given question = 40
Numbers of students born in August = 6
The probability that a student of the class was born in August = (Total numbers of students in the class) /
(Numbers of students born in August)
= 6/40 = 3/20

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

Solution:

Number of times 2 heads come up (in the given question) = 72
Total number of times the coins were tossed = 200
The probability of 2 heads coming up = (Number of times 2 heads come up) / (Total number of times the coins were tossed)
= 72/200 = 9/25

### The information gathered is listed in the table below:

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 (According to question)

(i) Numbers of families earning ₹10000 –13000 per month and owning exactly 2 vehicles = 29
The probability that the family chosen is earning ₹10000 – 13000 per month and owning exactly 2 vehicles =
(Numbers of families earning ₹10000 –13000 per month and owning exactly 2 vehicles) / (Total number of families)
= 29/2400

(ii) Number of families earning ₹16000 or more per month and owning exactly 1 vehicle = 579
The probability that the family chosen is earning ₹16000 or more per month and owning exactly 1 vehicle =
(Number of families earning ₹16000 or more per month and owning exactly 1 vehicle) / (Total number of families)
=579/2400

(iii) Number of families earning less than ₹7000 per month and does not own any vehicle = 10
The probability that the family chosen is earning less than ₹7000 per month and does not own any vehicle =
(Number of families earning less than ₹7000 per month and does not own any vehicle)/(Total number of families)
= 10/2400 = 1/240

(iv) Number of families earning ₹13000-16000 per month and owning more than 2 vehicles = 25
The probability that the family chosen is earning ₹13000 – 16000 per month and owning more than 2 vehicles =
(Number of families earning ₹13000-16000 per month and owning more than 2 vehicles ) / (Total number of families)
= 25/2400 = 1/96

(v) Number of families owning not more than 1 vehicle = 10+160+0+305+1+535+2+469+1+579 = 2062
The probability that the family chosen owns not more than 1 vehicle = 2062/2400 = 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:

Total number of students = 90
(given in question)
(i) Number of students who obtained less than 20% in the mathematics test = 7
The probability that a student obtained less than 20% in the mathematics test =
( Number of students who obtained less than 20% in the mathematics test)/(Total number of students)
= 7/90

(ii) Number of students who obtained marks 60 or above = 15+8 = 23

The probability that a student obtained marks 60 or above =

(Number of students who obtained marks 60 or above ) / (Total number of students)

= 23/90

### Find the probability that a student chosen at random

(i) likes statistics, (ii) does not like it.

Solution:

Total number of students = 135+65 = 200 (According to question)
(i) Number of students who like statistics = 135
The probability that a student likes statistics = (Number of students who like statistics) / (Total number of students)
= 135/200 = 27/40

(ii) Number of students who do not like statistics = 65
The probability that a student does not like statistics =
(Number of students who do not like statistics) / (Total number of students)
= 65/200 = 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 ½ km from her place of work?

Solution:

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

Total numbers of engineers = 40
(According to question)

(i) Number of engineers living less than 7 km from their place of work = 9

The probability that an engineer lives less than 7 km from her place of work =
(Number of engineers living less than 7 km from their place of work) / (Total numbers of engineers )
= 9/40

(ii) Number of engineers living more than or equal to 7 km from their place of work = 40-9 = 31

Probability that an engineer lives more than or equal to 7 km from her place of work =
(Number of engineers living more than or equal to 7 km from their place of work ) / (Total numbers of engineers)
= 31/40

(iii) Number of engineers living within ½ km from their place of work = 0

The probability that an engineer lives within ½ km from her place of work =
(Number of engineers living within ½ km from their place of work) / (Total numbers of engineers)
=0/40 = 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.

Solution:

The question is an activity to be performed by the students.

### 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:

The question is an activity to be performed by the students.

### Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Solution:

Data given in the question
Total number of bags present = 11
Number of bags containing more than 5 kg of flour = 7
The probability that any of the bags chosen at random contains more than 5 kg of flour =
(Number of bags containing more than 5 kg of flour) / (Total number of bags present)
= 7/11

### 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

Solution:

Total number of days in which the data was recorded = 30 days (According to the question)
Numbers of days in which sulphur dioxide was present in between the interval 0.12-0.16 = 2
The probability of the concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days =
(Numbers of days in which sulphur dioxide was present in between the interval 0.12-0.16) /
(Total number of days in which the data was recorded )
= 2/30 = 1/15

### 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.

Solution:

Total numbers of students = 30 (according to questions)
Number of students having blood group AB = 3
The probability that a student of this class, selected at random, has blood group

AB = (Number of students having blood group AB) / (Total numbers of students)
= 3/30 = 1/10

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