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Question

A bag contains 5 black, 4 green and 6 red balls. If a ball is drawn at random from the bag, find the probability that it will be:
i) red
ii) black
iii) green
[3 MARKS]


Solution

Each subpart: 1 Mark each

Total number of balls in the bag
= 5(B) + 4(G) + 6(R) = 15 balls
Thus, n(S) = 15
i) Event = {Red ball}, i.e., n(E) = 6
P(E)=n(E)n(S)=615=25

ii) Event = {Black ball}, i.e., n(E) = 5
P(E)=n(E)n(S)=515=13

iii) Event = {Green ball}, i.e., n(E) = 4
P(E)=n(E)n(S)=415

Mathematics

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