Question

A construction company has bid on two contracts. The probability of winning contract A is 0.35. Given the company wins contract A, the probability of winning contract B is 0.40. Given the company loses contract A, the probability of winning contract B decreases to 0.25. Given the information that the company won contract B, what is the probability that they also won contract A?

0.4628

0.5185

0.5895

0.4179

0.1400

Answer 1

Let P(A) = Probability of winning contract A. Given, P(A) = 0.35 ------- (1)

Let P(A') = Probability of not winning contract A. P(A') = 1 - P(A) = 0.65 ------ (2)

P( Winning B, given that A has been won) = P(B/A) = 0.4 ------- (3)

P( Winning B, given that A has not been won) = P(B/A') = 0.25 ------ (4)

To find : P( Winning A, given that B has been won) = P(A/B)

By Bayes Theorem: P(A / B) = P(A and B) / P(B)

From (3) P(B/A) = 0.4 = P(B and A) / P(A)

Therefore P(B and A) = 0.4 * P(A) = 0.4 * 0.35 = 0.14

Also P(B) = P(B/A) * P(A) + P(B/A') * P(A') = (0.4 * 0.35) + (0.25 * 0.65) = 0.14 + 0.1625 = 0.3025

Therefore :P(A / B) = P(A and B) / P(B) = 0.14 / 0.3025 = 0.4628 (Option 1)