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NCERT Solutions for Class 9 Maths Chapter 15 Probability
NCERT Solutions for Class 9 Maths Chapter 15 Probability are prepared by CBSE student eCARE expert to score good marks in class 9. This chapter contain many topics which are very important to score good in class 9. There are some important topic/Activity mentioned below:
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 digit is divisible by 3.
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?
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Some questions are asked by class 9 students related to Maths
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:
Here, the total number of trials = 30
? Number of times, the ball touched the boundary =6
∴ Number of times, the ball missed the boundary = 30 – 6 = 24
Let the event of not hitting the boundary be represented by E, then
P(E)=[No.oftimesthebatswomandidnothittheboundary][Totalnumberofballssheplayed]=2430=45
Thus, the required probability = 45
Question 2.
1500 families with 2 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 girl
Also, check whether the sum of these probabilities is 1.
Solution:
Here, total number of families = 1500
(i) ? Number of families having 2 girls = 475
∴ Probability of selecting a family having 2 girls = 4751500=1960
(ii) ? Number of families having 1 girl = 814
∴ Probability of selecting a family having 1 girl = 8141500=407750
(iii) Number of families having no girl = 211
Probability of selecting a family having no girl = 2111500
Now, the sum of the obtained probabilities
=1960+407750+2111500=475+814+2111500=15001500=1
i.e., Sum of the above probabilities is 1.