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?
(a) both contracts ;
(b) exactly one of the contracts ;
(c) neither of the contracts?
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Correct Answer: Option n
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) = \(P(A') \times P(B')\)
= \(0.5 \times 0.3\)
= \(0.15\)
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) = \(P(A') \times P(B')\)
= \(0.5 \times 0.3\)
= \(0.15\)