A bag contains 6 red balls, 8 blue balls and 10 yellow balls, all the balls being of the same size. If a ball is drawn from the bag, without looking into it, find the probability that the ball drawn is :

(i) yellow

(ii) red

(iii) blue

(iv) not yellow

(v) not blue

No. of possible outcomes = 24 (6 red + 8 blue + 10 yellow)

(i) No. of yellow balls = 10.

∴ No. of favourable outcomes = 10

P(drawing a yellow ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{10}{24} = \dfrac{5}{12}$.

Hence, probability of drawing a yellow ball = $\dfrac{5}{12}$.

(ii) No. of red balls = 6.

∴ No. of favourable outcomes = 6

P(drawing a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{24} = \dfrac{1}{4}$.

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

(iii) No. of blue balls = 8.

∴ No. of favourable outcomes = 8

P(drawing a blue ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{24} = \dfrac{1}{3}$.

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

(iv) No. of balls that are not yellow = 14 (6 red + 8 blue)

∴ No. of favourable outcomes = 14

P(not drawing a yellow ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{14}{24} = \dfrac{7}{12}$.

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

(v) No. of balls that are not blue = 16 (6 red + 10 yellow)

∴ No. of favourable outcomes = 16

P(not drawing a blue ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{16}{24} = \dfrac{2}{3}$.

Hence, probability of not drawing a blue ball = $\dfrac{2}{3}$.