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:
This question was reported to be attempted by majority of the candidates and they performed very well in it. It was one question where candidates were reported to have shown significant strength as majority of them scored full marks.
Candidates were reported to show that if the probability of winning contract X,P(X, then the probability of not winning contract X, P(X’) = 1 – P(X) = 1 - . Similarly, if the probability of not winning contract Y, P(Y’) the probability of winning contract Y, P(Y) = 1 – P(Y’) . Probability of winning both contracts was given by P(X) 0.7 Probability of winning exactly one contract = Probability of (winning X and not winning Y) or (not winning X and winning Y) = ( Probability of wining neither contract = Probability of not winning X and Y = .
, then the probability of not winning contract X, P(X’) = 1 – P(X) = 1 - 
. Similarly, if the probability of not winning contract Y, P(Y’)
the probability of winning contract Y, P(Y) = 1 – P(Y’) 
. Probability of winning both contracts was given by P(X) 
0.7
Probability of winning exactly one contract = Probability of (winning X and not winning Y) or (not winning X and winning Y) = (
Probability of wining neither contract = Probability of not winning X and Y = 
.