Question

(i) A box contains 80 discs, which are numbered from 1 to 80. If one disc is drawn at random from the box, find the probability that it bears a perfect square number. [CBSE 2011]
(ii) A box contains 90 discs which are numbered 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(a) a two-digit number
(b) a number divisible by 5 [CBSE 2017]

Answer 1

(i)
Total number of outcomes = 80.

Let E₁ be the event of getting a perfect square number.

Out of these numbers, perfect square numbers are 1, 4, 9, 16, 25, 36, 49 and 64.

Thus, number of favourable outcomes = 8.

∴ P(getting a perfect square number) = P(E₁) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_1}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{8}{80}=\frac{1}{10}$

Thus, the probability that the disc bears a perfect square number is $\frac{1}{10}$.

(ii)
Total possible outcomes = 90.
(a) Two-digit numbers are 10,11,12,13...90.
P(getting a two-digit number)=$\frac{81}{90}=\frac{9}{10}.$

(b) Numbers divisible by 5 are 5,10,15,20...90.
P(number divisible by 5)=$\frac{18}{90}=\frac{1}{5}.$