A bag contains 6 black, 5 white and 9 green balls. One ball is drawn at random. What is the probability that the ball drawn is:

(i) black?

(ii) not green?

(iii) either white or green?

(iv) neither white nor black?

Given,

Total number of outcomes = 6 (Black) + 5 (White) + 9 (Green) = 20

(i) Let A be the event of getting black ball, then

∴ The number of favourable outcomes to the event A = 6

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{20} = \dfrac{3}{10}$

Hence, the probability of getting black ball is $\dfrac{3}{10}$.

(ii) Let B be the event of not getting green ball, then

∴ The number of favourable outcomes to the event B = 6 (Black) + 5 (White) = 11

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{11}{20}$

Hence, the probability of not getting green ball is $\dfrac{11}{20}$.

(iii) Let C be the event of getting either white or green ball, then

∴ The number of favourable outcomes to the event C = 5 (White) + 9 (Green) = 14

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{14}{20}= \dfrac{7}{10}$

Hence, the probability of getting either white or green ball is $\dfrac{7}{10}$.

(iv) Let D be the event of getting neither white nor black ball , then

∴ The number of favourable outcomes to the event D = 9(green balls)

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{9}{20}$

Hence, the probability of getting neither white nor black ball is $\dfrac{9}{20}$.