A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be :

(i) not red ?

(ii) neither red nor green ?

(iii) white or green ?

Total number of possible outcomes = 10 + 16 + 8 = 34.

(i) No. of non red balls = 24 (16 white + 8 green)

∴ No. of favourable outcomes = 24.

P(drawing a not red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{\text{No. of possible outcomes}}} = \dfrac{24}{34} = \dfrac{12}{17}$.

Hence, the probability of drawing a not red ball = $\dfrac{12}{17}$.

(ii) Since, there are only 3 different colour balls in the bag.

∴ P(drawing neither red nor green ball) = P(drawing a white ball)

No. of favourable outcomes (of getting white ball) = 16.

P(drawing a white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{\text{No. of possible outcomes}}} = \dfrac{16}{34} = \dfrac{8}{17}$.

∴ P(drawing neither red nor green ball) = $\dfrac{8}{17}$.

Hence, the probability that the ball drawn is neither red nor green is $\dfrac{8}{17}$.

(iii) No. of white or green balls = 24 (16 white + 8 green)

∴ No. of favourable outcomes = 24.

P(drawing a white or green ball)

= $\dfrac{\text{No. of favourable outcomes}}{\text{\text{No. of possible outcomes}}} = \dfrac{24}{34} = \dfrac{12}{17}$.

Hence, the probability of drawing a white or green ball = $\dfrac{12}{17}$.