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) = `("Number of outcomes favourable to" E_2)/"Number of all possible outcomes"`

`= 13/65 = 1/5`                                                                     

Thus, the probability that the card bears a number divisible by 5 is `1/5`.