Question

A box contains cards bearing numbers 6 to 70. If one card is drawn at random from the box, find the probability that it bears a number divisible by 5.

Answer 1

​Given number 6, 7, 8, .... , 70 form an AP with a = 6 and d = 1.
Let T_n = 70. Then,
6 + (n − 1)1 = 70
⇒ 6 + n  − 1 = 70
⇒ n = 65

Thus, total number of outcomes = 65.

Let E₂ be the event of getting a number divisible by 5.

Out of these numbers, numbers divisible by 5 are 10, 15, 20, ... , 70.
Given number 10, 15, 20, .... , 70 form an AP with a = 10 and d = 5.
Let T_n = 70. Then,
10 + (n − 1)5 = 70
⇒ 10 + 5n  − 5 = 70
⇒ 5n = 65
⇒ n = 13

Thus, number of favourable outcomes = 13.

∴ P(getting a number divisible by 5) = P(E₂) = ${\displaystyle \frac{\text{Number of outcomes favourable to}{E}_2}{\text{Number of all possible outcomes}}}$

${\displaystyle =\frac{13}{65}=\frac{1}{5}}$                                                                      

Thus, the probability that the card bears a number divisible by 5 is ${\displaystyle \frac{1}{5}}$.