Answer:
Given:
Probability that the Yankees wins a game is
P(A) = 0.46
Probability that the Yankees loses a game is
P(A') = 1 - P(A') = 1 - 0.46 = 0.54
Probability that the Yankees scores 5 or more runs in a game is
P(B) = 0.59
Probability that the Yankees scores fewer than 5 runs in a game is
P(B') = 1 - P(B) = 1 - 0.59 = 0.41
Probability that the Yankees wins and scores 5 or more runs is
P(A⋂B) = 0.39
Applying the De Morgan's law, the probability that the Yankees scores fewer than 5 runs and they loses the game would satisfy:
1 - P(A'⋂B') = P(A) ⋃ P(B) = P(A) + P(B) - P(A⋂B)
or
1 - P(A'⋂B') = 0.46 + 0.59 - 0.39
or
1 - P(A'⋂B') = 0.66
=> P(A'⋂B') = 1 - 0.66 = 0.34
Applying the Bayes theorem, the probability that the Yankees would score fewer than 5 runs, given they lose the game:
P(B'|A') = P(A'⋂B')/P(A')
or
P(B'|A') = 0.34/0.54 = 0.630
Hope this helps!
:)
