Question

Cards numbered 1 to 30 are put in a bag. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) not divisible by 3,
(ii) a prime number greater than 7,
(iii) not a perfect square number.

Answer 1

Total number of outcomes = 30.

(i) ​Let E₁ be the event of getting a number not divisible by 3.

Out of these numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30.

Number of favourable outcomes = 30 − 10 = 20

∴ P(getting a number not divisible by 3) = P(E₁) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_1}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{20}{30}=\frac{2}{3}$

Thus, the probability that the number on the card is not divisible by 3 is $\frac{2}{3}$.

(ii) ​Let E₂ be the event of getting a prime number greater than 7.

Out of these numbers, prime numbers greater than 7 are 11, 13, 17, 19, 23 and 29.

Number of favourable outcomes = 6

∴ P(getting a prime number greater than 7) = P(E₂) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_2}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{6}{30}=\frac{1}{5}$

Thus, the probability that the number on the card is a prime number greater than 7 is $\frac{1}{5}$.

(iii) ​Let E₃ be the event of getting a number which is not a perfect square number.

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

Number of favourable outcomes = 30 − 5 = 25

∴ P(getting non-perfect square number) = P(E₃) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_3}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{25}{30}=\frac{5}{6}$

Thus, the probability that the number on the card is not a perfect square number is $\frac{5}{6}$.