Chapter 16 Probability Ex 16.3 Solutions
Question - 21 : - In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that
(i) The student opted for NCC or NSS.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.
Answer - 21 : -
From the question it is given that,
The total number of students in class = 60
Thus, the sample space consist of n(S) = 60
Let us assume that the students opted for NCC be ‘A’
And also assume that the students opted for NSS be ‘B’
So, n(A) = 30, n(B) = 32 , n(A∩B) = 24
We know that, P(A) = n(A)/n(S)
= 30/60
= ½
P(B) = n(B)/n(S)
= 32/60
= 8/15
P(A∩B) = n(A∩B)/n(S)
= 24/60
= 2/5
Therefore, P(A∪B)= P(A) + P(B) – P(A∩B)
(i) The student opted for NCC or NSS.
P (A or B) = P(A) + P(B) –P(A and B)
P(A∪B) = P(A) + P(B) – P(A∩B)
= ½ + (8/15) – (2/5)
= 19/30
(ii) P(student opted neither NCC nor NSS)
P(not A and not B) = P(AI∩BI)
We know that, P(AI∩BI) = 1 – P(A∪B)
= 1 – (19/30)
= 11/30
(iii) P(student opted NSS but not NCC)
n(B – A) = n(B) – n (A∩B)
⇒ 32 – 24 = 8
The probability that the selected student has opted for NSSand not NCC is
= (8/60) = 2/15