Q50 / 90 · JEE Main 2019 (12 Jan Shift 2) · Mathematics

In a class of $60$ students, $40$ opted for NCC, $30$ opted for NSS and $20$ opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is :

Solution

Let $A & B$ be the sets for NCC and NSS respectively.

Given $n (A) = 40, n (B) = 30, n (A \cap B) = 20$

$∴ n (A \cup B) = n (A) + n (B) - n (A \cap B)$

$= 40 + 30 - 20 = 50$

$∴ P (A \cup B) = \frac{50}{60} = \frac{5}{6}$

$∴ P {(A \cup B)}^{'} = 1 - P (A \cup B)$

$= 1 - \frac{5}{6} = \frac{1}{6}$

Hence, the correct answer is option: 'A'.