NCERT Solutions for Class 9 Maths Chapter 15 Probability

NCERT Solutions for Class 9 Maths Chapter 15 Probability is an article that contains all the resources for learning the solution to the problems given in the CBSE syllabus 2023-24. Using this NCERT Solutions, students can learn the methods for the solution of problems in a step-by-step manner. In NCERT Solutions for Class 9 Maths, all the problems given in the NCERT book for Class 9 Chapter 15 are solved. These Probability Class 9 NCERT Book Solutions are the perfect resource to help you master the chapter thoroughly.

Probability explores the idea of the likelihood that a particular experiment outcome will occur. Only one task, based on issues from real-world instances, is provided in this chapter to help the reader comprehend the experimental approach to probability. Events and the empirical formula for probability are some topics discussed in the article.

Class 9 Maths NCERT Solutions Chapter 15 Exercises
NCERT Maths Solutions Class 9 Exercise 15.1 – 10 Questions (4 Short Answers, 3 Long Answers, 3 Very Long Answers)

Discover what's covered in the solutions for Chapter 15 Maths Class 9

Exercises provided in the NCERT Solutions for Class 9 Maths Chapter 15 Probability

NCERT Solutions for Class 9 Maths Chapter 15, titled 'Probability,' delves into the concept of probability and its versatile applications across different domains. This chapter encompasses the following exercises:

NCERT Solutions for Class 9 Mathematics Chapter 15: Exercise 15.1

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:

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 subtract 
⇒(Total number of balls batswoman plays) - (Numbers of boundary hit by batswoman) 
⇒ 30 – 6 = 24 
Probability of that she didn't hit a boundary = 24/30 = 4/5

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

Solution:

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

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

(iii) Numbers of families having 2 girls = 211 
Probability of chosen 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

Question 4. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

Outcome	3heads	2heads	1 head	No heads
Frequency	23	72	77	28

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

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 ₹)	0	1	2	Above 2
Less than 7000	10	160	25	0
7000-10000	0	305	27	2
7000-10000	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 (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

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 = 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.

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:

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

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.

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

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.

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

Things to Keep in Mind Before Working on NCERT Class 9 Maths Chapter 15 - Probability

• All the solutions in the article are provided by the team of professionals at Gfg, which help users learn solutions with ease.
• These solutions are accurate and comprehensive in nature, which made them very simple and easy to understand.
• All the provided solutions are discussed in a step-by-step manner which helps students learn solutions with all of its intermediate steps.

Types of Questions Asked in Probability Class 9 Maths NCERT 

Here is a list of the different types of questions you'll encounter in Chapter 15 of NCERT Maths for Class 9:

1. Questions related to coins.
2. Questions related to rolling dice.
3. Questions involving a pack of cards.
4. Questions related to the number system, like finding the probability of digits being even, odd, divisible by 3 or 4, etc.
5. Questions about the occurrence of events.

Probability - Class 9 Math Chapter 15 NCERT Solutions Summary

Class 9 marks a significant transition for students as they step into the world of board exams. This can be both exciting and a bit daunting. The syllabus expands, introducing new and challenging subjects, especially mathematics, with unfamiliar concepts. Mastering these chapters and concepts is vital for building a strong foundation. The use of Class 9 Probability NCERT Solutions can greatly enhance your understanding, boosting your confidence as you progress to higher classes. These NCERT Solutions for Class 9 Probability serve as the building blocks of your knowledge, strengthening your educational base for future endeavors.

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