Question -
Answer -
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