# Mark the correct alternative in each of the following:

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]

Solution:

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.

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

Hence, the correct answer is option C.