Question

A bag contains 4 black, 6 white and 8 red balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is:
i) white
ii) not red
iii) red or white
iv) red and white
v) neither black nor white
vi) either red or black
[6 MARKS]

Answer 1

Each subpart: 1 Mark each

Total number of balls = 4 + 6 + 8 = 18
i) Number of white balls in the bag $=6$
P(white) $=\frac{6}{18}=\frac{1}{3}$

ii) Number of balls which are not red $=18-8=10$
P(not red) $=\frac{10}{18}=\frac{5}{9}$

iii) Number of balls which are either red or white $=8+6=14$
P(red or white) $=\frac{14}{18}=\frac{7}{9}$

iv) There are no balls which are both red and white.
P(red and white) $=\frac{0}{18}=0$ (Impossible event)

v) Number of balls which are neither black nor white $=18-4-6=8$
P(neither black nor white) $=\frac{8}{18}=\frac{4}{9}$

vi) Number of balls which are either red or black $=8+4=12$
P(either red or black) $=\frac{12}{18}=\frac{2}{3}$