A box contains a certain number of balls. On each of 60% balls, letter A is marked. On each of 30% balls, letter B is marked and on each of remaining balls, letter C is marked. A ball is drawn from the box at random. Find the probability that the ball drawn is :

(i) marked C

(ii) A or B

(iii) neither B nor C.

Given,

On each of 60% balls, letter A is marked. On each of 30% balls, letter B is marked and on each of remaining balls (i.e. 10%), letter C is marked.

Let no. of balls be x.

∴ No. of possible outcomes = x.

No. of balls marked A = $\dfrac{60}{100} \times x = \dfrac{3x}{5}$

No. of balls marked B = $\dfrac{30}{100} \times x = \dfrac{3x}{10}$

No. of balls marked C = $\dfrac{10}{100} \times x = \dfrac{x}{10}$.

(i) No. of balls marked C or favourable outcomes = $\dfrac{x}{10}$

P(drawing a ball marked C) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{\dfrac{x}{10}}{x} = \dfrac{1}{10}$.

Hence, the probability of drawing a ball marked C = $\dfrac{1}{10}$.

(ii) No. of balls marked A or B = $\dfrac{3x}{5} + \dfrac{3x}{10} = \dfrac{6x + 3x}{10} = \dfrac{9x}{10}$.

P(drawing a ball marked A or B) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{\dfrac{9x}{10}}{x} = \dfrac{9}{10}$.

Hence, the probability of drawing a ball marked A or B = $\dfrac{9}{10}$.

(iii) Since, the balls are marked either A, B or C.

So, P(drawing neither B nor C) = P(drawing A marked ball)

No. of A marked balls = $\dfrac{3x}{5}$.

∴ No. of favourable outcomes = $\dfrac{3x}{5}$.

P(drawing a ball marked A) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{\dfrac{3x}{5}}{x} = \dfrac{3}{5}$.

∴ P(drawing neither B nor C) = $\dfrac{3}{5}$.

Hence, the probability of drawing ball marked neither B nor C = $\dfrac{3}{5}$.