Question

A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears prime number less than 23 is
(a) $\frac{7}{90}$ (b) $\frac{1}{9}$ (c) $\frac{4}{45}$ (d) $\frac{8}{89}$

Answer 1

Total number of discs numbered 1 to 90 : 90

Prime numbers less than 23 are :2, 3, 5, 7, 11, 13, 17, 19

Number of favourable outcomes = 8

$\text{Probability (getting disc of prime number less than 23)}=\frac{\text{Number of favourable outcomes}}{\text{Total outcomes}}$

P(getting prime number less than 23) $=\frac{8}{90}=\frac{4}{45}$

answer:- (c) $\frac{4}{45}$