Question

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 1

Let, A be the event such that selected student opted for NCC and B the event for selected student has opted for NSS.

Total number of students is 60, 30 students opted NCC, 32 opted NSS and 24 opted both NSS and NCC.

The probability of students opted NCC,

P( A ) = $\frac{30}{60}$ = $\frac{1}{2}$

The probability of students opted NSS,

P( B )= $\frac{32}{60}$ = $\frac{8}{15}$

The probability of students opted both NSS and NCC,

P( A∩B ) = $\frac{24}{60}$ = $\frac{2}{5}$

(i)

The probability of students opted both NSS or NCC,

P( A∪B ) = P( A ) + P( B ) − P( A∩B ) = $\frac{1}{2}$ + $\frac{8}{15}$ − $\frac{2}{5}$ = $\frac{19}{20}$

(ii)

The probability of students opted neither NSS nor NCC,

P( A ′ ∩ B ′) = P (A∪B) ′ = 1 − P( A∪B )

= 1 − $\frac{19}{20}$

= $\frac{11}{20}$

(iii)

Number of students who have opted for NSS but not NCC is,

n( B−A ) = n(B ) − n( A∩B ) = 32 − 24 = 8

The probability}{ for selected student has opted for NSS but not for NCC is,

$\frac{8}{60}$ = $\frac{2}{15}$