A building contractor tendered for two independent contracts, X and Y. The probabilities that he will win contract X is 0.5 and not win contract Y is 0.3, What is the probability that he will win :
(a) both contracts ;
(b) exactly one of the contracts ;
(c) neither of the contracts?
Explanation
Let A and B denote the events that the man wins contracts X and Y respectively.
Then P(A) = 0.5
P(A’) = 1 – 0.5 = 0.5
P(B’) = 0.3
P(B) = 1 – 0.3 = 0.7
(a) The probability that the man wins both contracts = (0.5 times 0.7 = 0.35).
(b) The probability that the man wins exactly one of the contracts is (P(A) times P(B’) + P(B) times P(A’))
= (0.5 times 0.3 + 0.7 times 0.5)
= (0.15 + 0.35)
= (0.50)
(c) Neither of the contracts (i.e not X, not Y) = (P(A’) times P(B’))
= (0.5 times 0.3)
= (0.15)