Question

Cards marked with numbers 1, 3, 5, ..., 101 are placed in a bag and mixed thoroughly. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) less than 19,
(ii) a prime number less than 20.

Answer 1

​Given number 1, 3, 5, ..., 101 form an AP with a = 1 and d = 2.
Let T_n = 101. Then,
1 + (n − 1)2 = 101
⇒ 1 + 2n − 2 = 101
⇒ 2n = 102
⇒ n = 51

Thus, total number of outcomes = 51.

(i) Let E₁ be the event of getting a number less than 19.

Out of these numbers, numbers less than 19 are 1, 3, 5, ... , 17.
Given number 1, 3, 5, .... , 17 form an AP with a = 1 and d = 2.
Let T_n = 17. Then,
1 + (n − 1)2 = 17
⇒ 1 + 2n − 2 = 17
⇒ 2n = 18
⇒ n = 9

Thus, number of favourable outcomes = 9.

∴ P(getting a number less than 19) = P(E₁) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_1}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{9}{51}=\frac{3}{17}$

Thus, the probability that the number on the drawn card is less than 19 is $\frac{3}{17}$.

(ii) Let E₂ be the event of getting a prime number less than 20.

Out of these numbers, prime numbers less than 20 are 3, 5, 7, 11, 13, 17 and 19.

Thus, number of favourable outcomes = 7.

∴ P(getting a prime number less than 20) = P(E₂) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_2}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{7}{51}$

Thus, the probability that the number on the drawn card is a prime number less than 20 is $\frac{7}{51}$.