Question

A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls. A ball is drawn at random from the bag. Find the probability that it is

(i) black,
(ii) not green,
(iii) red or white,
(iv) neither red nor green.

Answer 1

Total number of balls = 15

(i) Number of black balls = 2

∴ P(getting a black ball) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{2}{15}$

Thus, the probability of getting a black ball is $\frac{2}{15}$.

(ii) Number of balls which are not green = 4 + 5 + 2 = 11

∴ P(getting a ball which is not green) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{11}{15}$

Thus, the probability of getting a ball which is not green is $\frac{11}{15}$.

(iii) Number of balls which are either red or white = 4 + 5 = 9

∴ P(getting a ball which is red or white) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{9}{15}=\frac{3}{5}$

Thus, the probability of getting a ball which is red or white is $\frac{3}{5}$.

(iv) Number of balls which are neither red nor green = 4 + 2 = 6

∴ P(getting a ball which is neither red nor green) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{6}{15}=\frac{2}{5}$

Thus, the probability of getting a ball which is neither red nor green is $\frac{2}{5}$.