Question

Mark the correct alternative in each of the following:

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 a prime number less than 23, is

(a) $\frac{7}{90}$ (b) $\frac{10}{90}$ (c) $\frac{4}{45}$ (d) $\frac{9}{89}$ [CBSE 2013]

Answer 1

There are 90 discs numbered from 1 to 90.

∴ Total number of outcomes = 90

The prime numbers less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19.

So, the favourable number of outcomes are 8.

∴ P(disc drawn bears a prime number less than 23) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{8}{90}=\frac{4}{45}$

Hence, the correct answer is option C.