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 for neither NCC nor NSS.
(iii) The student has opted for NSS but not NCC.
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Solution:-
From the question, it is given that
The total number of students in the class = 60
Thus, the sample space consists of n(S) = 60
Let us assume that the students opted for NCC to be ‘A’.
And also assume that the students opted for NSS to 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 for 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 for NSS but not NCC)
n(B – A) = n(B) – n (A∩B)
⇒ 32 – 24 = 8
The probability that the selected student has opted for NSS and not NCC is
= (8/60) = 2/15