A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls. One ball is drawn at random from the bag. Find the probability that the ball is :

(i) white

(ii) red or black

(iii) not green

(iv) neither white nor black.

(i) No. of white balls = 8 and total no. of balls = 6 + 8 + 5 + 3 = 22.

Let E₁ be the event of choosing a white ball, then number of favourable outcomes to E₁ = 8

$P(E$1) = \dfrac{\text{No. of favourable outcomes to } E1}{\text{Total no. of possible outcomes}} = \dfrac{8}{22} = \dfrac{4}{11}.

Hence, the probability of drawing a white ball is $\dfrac{4}{11}$.

(ii) Total no. of red and black balls = 6 + 3 = 9.

Let E₂ be the event of choosing a red or black ball, then number of favourable outcomes to E₂ = 9

$P(E$2) = \dfrac{\text{No. of favourable outcomes to } E2}{\text{Total no. of possible outcomes}} = \dfrac{9}{22}.

Hence, the probability of drawing a red or black ball is $\dfrac{9}{22}$.

(iii) Probability of not drawing a green ball means probability of drawing any other colour ball.

Total no. of red, black and white balls = 6 + 3 + 8 = 17.

Let E₃ be the event of choosing a red, black or white ball, then number of favourable outcomes to E₃ = 17

$P(E$3) = \dfrac{\text{No. of favourable outcomes to } E3}{\text{Total no. of possible outcomes}} = \dfrac{17}{22}.

Hence, the probability of drawing a not green ball is $\dfrac{17}{22}$.

(iv) Probability of not drawing a white or black ball means probability of drawing red or green ball.

Let E₄ be the event of choosing a red or green ball, then number of favourable outcomes to E₄ = 11

$P(E$4) = \dfrac{\text{No. of favourable outcomes to } E4}{\text{Total no. of possible outcomes}} = \dfrac{11}{22} = \dfrac{1}{2}.

Hence, the probability of drawing neither white nor black ball is $\dfrac{1}{2}$.