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.