A ball is drawn at random from a box containing 12 white, 16 red and 20 green balls. Determine the probability that the ball drawn is :

(i) white

(ii) red

(iii) not green

(iv) red or white

Box contains 12 white, 16 red and 20 green balls.

∴ No. of possible outcomes = 48.

(i) There are 12 white balls.

∴ No. of favourable outcomes = 12

P(drawing a white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{12}{48} = \dfrac{1}{4}$.

Hence, the probability of drawing a white ball = $\dfrac{1}{4}$.

(ii) There are 16 red balls.

∴ No. of favourable outcomes = 16

P(drawing a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{16}{48} = \dfrac{1}{3}$.

Hence, the probability of drawing a red ball = $\dfrac{1}{3}$.

(iii) There are 28 non green balls (12 white + 16 red).

∴ No. of favourable outcomes = 28

P(not drawing a green ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{28}{48} = \dfrac{7}{12}$.

Hence, the probability of not drawing a green ball = $\dfrac{7}{12}$.

(iv) Since, there are only 3 different colour balls.

We can say that,

P(drawing a red or white ball) = P(not drawing a green ball) = $\dfrac{7}{12}$.

Hence, the probability of drawing a red or white ball = $\dfrac{7}{12}$.