A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is :

(i) a black ball.

(ii) a red ball.

(iii) a white ball.

(iv) not a red ball.

(v) not a black ball.

Total number of balls = 3 + 5 + 2 = 10

So, the total number of possible outcomes = 10

(i) There are 5 black balls.

∴ Number of favourable outcomes = 5

P(getting a black ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{10} = \dfrac{1}{2}$.

Hence, probability of getting a black ball = $\dfrac{1}{2}$.

(ii) There are 2 red balls.

∴ Number of favourable outcomes = 2.

P(getting a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{10} = \dfrac{1}{5}$.

Hence, probability of getting a red ball = $\dfrac{1}{5}$.

(iii) There are 3 white balls.

∴ Number of favourable outcomes = 3.

P(getting a white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{10}$.

Hence, probability of getting a white ball = $\dfrac{3}{10}$.

(iv) There are 2 red balls.

∴ 8 (10 - 2) balls which are not red.

∴ Number of favourable outcomes = 8

Thus, P(not getting a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{10} = \dfrac{4}{5}$.

Hence, the probability of not getting a red ball = $\dfrac{4}{5}$.

(v) There are 3 white + 2 red = 5 balls which are not black

∴ Number of favourable outcomes = 5

Thus, P(not getting a black ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{10} = \dfrac{1}{2}$.

Hence, the probability of not getting a black ball = $\dfrac{1}{2}$.