The probability of drawing an even number or number greater than 4 containing the numbers 1-10, inclusive.

would the probability of this event be:

a) 0.25
b) 0.5
c) 0.8
d) 1

Key word here, "or"...so you would add the unique probabilities of either together...

P(e)=1/2, P(>4)=3/10  (the numerator is three rather than five because 2 and 4 were already accounted for in P(e)...

P=5/10+3/10=8/10=4/5  (80%)

P=0.8

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!

:)