NCERT Solutions for Class 9 Maths Chapter 15 Probability
NCERT solutions for class 9 maths chapter 15 Probability: Predictions are common in our life. Whether it will rain today, whether India will win the match etc. Predictions are based on experiments and observation. Solutions of NCERT for class 9 maths chapter 15 Probability is covering the solutions to these kinds of reallife predictions. Suppose you are doing an experiment of tossing a coin 20 times then each toss is known as a trial and the possible outcomes of a toss are head and tail. If in a toss a tail has occurred we call that an event 'tail has occurred' and vice versa. In these CBSE NCERT solutions for class 9 maths chapter 15 Probability is designed to provide you step by step solutions to all such problems. In this chapter, there is only one exercise that consists of a total of 13 questions including some activities. NCERT solutions for class 9 maths chapter 15 Probability is covering every practice problem to help you while preparing for the final examinations. Apart from probability, you can access the free NCERT solutions by the given link.
NCERT solutions for class 9 maths chapter 15 Probability Excercise: 15.1
Answer:
From the above question, the data of interest is,
Total Number of balls batswoman played = 30
Number of times batswoman hits a boundary =6
Therefore, we can say,
Number of times batswoman could not hit a boundary =24
P(she did not hit a boundary)
=24/30 =0.80
Ans:0.80
Number of girls in a family  2  1  0 
Number of families  475  814  211 
Answer:
From the above question, the data that we can take is,
Total Number of families= 475+814+211 = 1500
Number of families having 2 girls in the family =475
We know:
Empirical (or experimental) probability P(E) of this event can be written as
P(Family, chosen at random has 2 girls) =
= 475/1500 = 19/60
Number of girls in a family  2  1  0 
Number of families  475  814  211 
Answer:
From the above question, the data that we can take is,
Total Number of families= 475+814+211 = 1500
Number of families having 1 girl in the family =814
We know:
Empirical (or experimental) probability P(E) of this event can be written as
P(Family, chosen at random has 1 girl) =
= 814/1500 = 407/1500
Number of girls in a family  2  1  0 
Number of families  475  814  211 
Answer:
From the above question, the data that we can take is,
Total Number of families= 475+814+211 = 1500
Number of families having no girls in the family =211
We know:
Empirical (or experimental) probability P(E) of this event can be written as
P(Family, chosen at random has no girls) =
= 211/1500
Ans:= 211/1500
Answer:
Total no. of students = 40
Total no. of students who all are born in August =6
P( a student of the class was born in August) =6/40
=3/20
Ans : 3/20
Outcome 
3 Heads 
2 Heads 
1 Head 
0 Head 
Frequencies 
23 
72 
77 
28 
Answer
Total number of times coins tossed = 200
Total number of possible outcomes = 72
Required probability = 72/200 => 9/25
Monthly income (in Rs)  Vehicles per family  
0  1  2  Above 2  
Less than 7000  10  160  25  0 
700010000  0  305  27  2 
1000013000  1  535  29  1 
1300016000  2  469  59  25 
16000 or more  1  579  82  88 
Answer:
Although it is given that,
Total no. of families= 2400
Let us find this by adding all the cases
= 10+0+1+2+1+160+305+535+469+579+25+27+29+59+82+0+2+1+25+88 =2400
Earning Rs. 10000 – Rs.13000 per month and owning exactly 2 vehicles =29
P(Earning Rs. 10000–Rs.13000 per month and owning exactly 2 vehicles)= P _{ 1 }
P _{ 1 } = 29/2400
Monthly income (in Rs)  Vehicles per family  
0  1  2  Above 2  
Less than 7000  10  160  25  0 
700010000  0  305  27  2 
1000013000  1  535  29  1 
1300016000  2  469  59  25 
16000 or more  1  579  82  88 
Answer:
Although it is given that,
Total no. of families= 2400
Let us find by this adding all the cases
= 10+0+1+2+1+160+305+535+469+579+25+27+29+59+82+0+2+1+25+88 =2400
Total no. of families earning Rs.16000 or more per month and owning exactly 1 vehicle = 579
P(earning Rs.16000 or more per month and owning exactly 1 vehicle )= P2
P2 = 579/2400
Monthly income (in Rs)  Vehicles per family  
0  1  2  Above 2  
Less than 7000  10  160  25  0 
700010000  0  305  27  2 
1000013000  1  535  29  1 
1300016000  2  469  59  25 
16000 or more  1  579  82  88 
Answer:
Although it is given that,
Total no. of families= 2400
Let us find by this adding all the cases
= 10+0+1+2+1+160+305+535+469+579+25+27+29+59+82+0+2+1+25+88 =2400
Total no. of families, those are earning less than Rs. 7000 per month and not having any vehicle = 10
P(earning less than Rs. 7000 per month and does not have any vehicle)= P3
P3 = 10/2400
=1/240
Ans: 1/240
Monthly income (in Rs)  Vehicles per family  
0  1  2  Above 2  
Less than 7000  10  160  25  0 
700010000  0  305  27  2 
1000013000  1  535  29  1 
1300016000  2  469  59  25 
16000 or more  1  579  82  88 
Answer:
Although it is given that,
Total no. of families= 2400
Let us find by this adding all the cases
= 10+0+1+2+1+160+305+535+469+579+25+27+29+59+82+0+2+1+25+88 =2400
Total number of families,those are earning Rs.13000 – 16000 per month and owning more than 2 vehicles = 25
P(earning Rs.13000 – 16000 per month and owning more than 2 vehicles)= P4
P4 = 25/2400
=1/96
Ans :1/96
Monthly income (in Rs)  Vehicles per family  
0  1  2  Above 2  
Less than 7000  10  160  25  0 
700010000  0  305  27  2 
1000013000  1  535  29  1 
1300016000  2  469  59  25 
16000 or more  1  579  82  88 
Answer:
Although it is given that,
Total no. of families= 2400
Let us find by this adding all the cases
= 10+0+1+2+1+160+305+535+469+579+25+27+29+59+82+0+2+1+25+88 =2400
Total number of families owning not more than 1 vehicle= 10+0+1+2+1+160+305+535+469+579 = 2062
Let, P(families owning not more than 1 vehicle)= P5
P5 = 2062/2400
Answer:
Marks  Number of students 
020  2 
2030  10 
3040  10 
4050  20 
5060  20 
6070  15 
70 above  8 
Total  90 
From the above question, the data of our interest is:
Total no. of students =90
Total no. of students who obtained less than 20% in the mathematics test= 7
P(student obtained less than 20% in the mathematics test) = 7/90
Ans: 7/90
Q6 (ii) Refer to Table 14.7, Chapter 14. Find the probability that a student obtained marks 60 or above.
Answer:
Marks  Number of students 
020  2 
2030  10 
3040  10 
4050  20 
5060  20 
6070  15 
70 above  8 
Total  90 
From the above question, the data of our interest is:
Total no. of students =90
Total no. of students who obtained marks 60 or above = 15+8 =23
P(a student obtains marks 60 or above) = 23/90
Ans: 23/90
Find the probability that a student is chosen at random likes statistics,
Answer:
From the above question, the data of our interest is:
Total no. of students =135+65 =200
Total no. of students who like statistics = 135
P(students like statistics )= 135/200 =27/40
Ans: 27/40
Find the probability that a student chosen at random does not like it.
Answer:
From the above question, the data of our interest is:
Total no. of students =135+65=200
Total no. of students who do not like it.= 65
P(a student does not like it) = 65/200
= 13/40
Ans: 13/40
Answer:
The distance (in km) of 40 engineers from their residene 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 
Total no. of engineers = 40
Total no. of engineers who are living less than 7 km from their workplace = 9
Therefore we can say,
P(engineers who are living less than 7 km from their workplace)=
= 9/40
Ans: 9/40
Answer:
The distance (in km) of 40 engineers from their residene 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 
Total no. of engineers = 40
Total no. of engineers who are living less than 7 km from their workplace = 31
Therefore we can say,
P(engineers who are living more than or equal to 7 km from their workplace)=31/40
Ans: 31/40
Answer:
The distance (in km) of 40 engineers from their residene 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 
Hey, don't you think its too simple:
Well, there is no such engineer whose distance between residence and place of work is less than 1/2 km
Therefore,
engineer whose distance between residence and place of work is less than 1/2 km =0
P(engineer whose distance between residence and place of work is less than 1/2 km) = 0
Ans: 0
Answer:
This activity can be taken as a general problem:
Assumption:
Let the frequency of twowheelers = x
Let the frequency of threewheelers = y
Let the frequency of fourwheelers = z
Total no. of vehicles= x+y+z
therefore,
P(anyone vehicle out of the total vehicles I have observed is a twowheeler) =
Answer:
This activity can be taken as a general problem:
Well, we know the divisibility by 3 is when the sum of all the digits is divisible by 3
So,
The student will write the number between 100999
There are 900 3  digit numbers , which are 100, 101, 102, 103, ..., 999.
The first 3digit numbers that are exactly divisible by 3 is 102, 105, ..... 999
total numbers which are divisible by 3 = 300
P(the sum of all the digits is divisible by 3) = 300/900
=1/3
Ans: 1/3
Answer:
From the above question, the data of our interest is:
Total no. of bags = 11
Total no. of bags that contain more than 5 kg flour = 7
P(Bag contain more than 5 kg flour) = 7/11
Ans: 7/11
Answer:
The data is representing the concentration of sulfur dioxide in the air in parts per million (ppm) of a city. The data obtained for a month of 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 
Total no. of days =30
No. of days in which Conc. of sulfur dioxide in the interval 0.12  0.16 = 2
P(Conc. of sulphur dioxide in the interval 0.12  0.16) =2/30
Ans: 1/15
Answer:
The belowwritten data is representing the blood groups of 30 students study in class VIII.
A, B, O, O, B, A, O, O, AB, O, A, O, B, A, O
A, AB, O, A, A, B, A, B, O, O, O, AB, B, A, O
Total no. of students = 30
Total no. of students of this class who has blood group AB =3
P(Student of this class has blood group AB)= 3/30 => 1/10
Ans: 1/10
NCERT solutions for class 9 maths chapter wise
Chapter No. 
Chapter Name 
Chapter 1 

Chapter 2 
CBSE NCERT solutions for class 9 maths chapter 2 Polynomials 
Chapter 3 
Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry 
Chapter 4 
NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables 
Chapter 5 
CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry 
Chapter 6 

Chapter 7 

Chapter 8 
CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals 
Chapter 9 
Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles 
Chapter 10 

Chapter 11 
CBSE NCERT solutions for class 9 maths chapter 11 Constructions 
Chapter 12 

Chapter 13 
NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes 
Chapter 14 
CBSE NCERT solutions for class 9 maths chapter 14 Statistics 
Chapter 15 
NCERT solutions for class 9 maths chapter 15 Probability 
NCERT solutions for class 9 subject wise
How to use NCERT solutions for class 9 maths chapter 15 Probability?
 Try to understand the concept of this chapter using the theory given in the NCERT textbook.
 Connect the concepts to reallife problems.
 Go through some examples to understand the way of solving a problem.
 Practice the questions given in the practice exercises.
 During the practice, you can use NCERT solutions for class 9 maths chapter 15 Probability as an assistant.