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.
(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

Answer 1

(i) The 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, the number of favorable outcomes = 8.

Therefore,
P(getting a perfect square number) = P(E1) = Number of outcomes favorable toE1 / Number of all possible outcomes
= 8/80= 1/10

Thus, the probability that the disc bears a perfect square number is 1/10.

(ii)

Total number of discs = 90

(a) Total number of two-digit numbers between 1 and 90 = 81

P (getting a two-digit number) = $\frac{81}{90}$ =$\frac{9}{10}$

(b) Numbers that are between 1 and 90 and divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, and 90. Therefore, total numbers divisible by 5 = 18

Probability of getting a number divisible by 5= $\frac{18}{90}$ = $\frac{1}{5}$