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?
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?

Solution:

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

So, n(S) = 34

(i) Favorable outcomes for not a red ball = favorable outcomes for white or green ball

So, number of favorable outcomes for white or green ball = 16 + 8 =24 = n(E)

Hence, probability for not drawing a red ball = n(E)/ n(S) = 24/34 = 12/17

(ii) Favorable outcomes for neither a red nor a green ball = favorable outcomes for white ball

So, the number of favorable outcomes for white ball = 16 = n(E)

Hence, probability for not drawing a red or green ball = n(E)/ n(S) = 16/34 = 8/17