Question

A bag contains 8 red, 6 white and 4 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is 

(i) red or white 

(ii) not black 

(iii) neither white or black

Answer 1

Total number of possible outcomes, n(S) = 18 

(i) Number of favorable outcomes, 

n(E) = 14

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{14}{18}\) = \(\frac{7}{9}\)

(ii) Number of events of getting a black ball, 

n(E) = 4

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{18}\) = \(\frac{2}{9}\)

Probability of not getting a black ball = 1 – P(E)

= 1 - \(\frac{2}{9}\) = \(\frac{7}{9}\)

(iii) Number of favorable outcomes, 

n(E) = 8

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{8}{18}\) = \(\frac{4}{9}\)