Question

A bag contains 5 red, 4 blue and 3 green balls. One ball is drawn at random. Find the probability that the ball drawn is
i) blue
ii) not red
iii) either red or green
iv) black
[4 MARKS]

Answer 1

Each subpart: 1 Mark each

Total number of balls = 5 + 4 + 3 = 12
i) Number of favourable outcomes = 4
$\therefore$ P(blue) $=\frac{4}{12}=\frac{1}{3}$

ii) P(not red) = 1 - P(red) $=1-\frac{5}{12}=\frac{7}{12}$

iii) For either red or green balls, we have favourable outcomes = 5 + 3 = 8
$\therefore$ P(either red or green) $=\frac{8}{12}=\frac{2}{3}$

iv) There is no black ball in the bag, thus P(black) = 0 (impossible event)