The probability that Marvin will win in a certain game whenever he plays is . If he plays twice, find the probability that he will win just once.
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He plays twice (two independent acts). He wins just once if
a. he wins the first game and loses the second game; or, if
b. he loses the first game and wins the second game
The probability that Marvin will lose in any game is 3/4. Hence,
P(a) = 1 / 4 ∗ 3 / 4 = 3 / 1 6
P(b) = 3 / 4 ∗ 1 / 4 = 3 / 1 6
P(a) or P(b) = 3 / 1 6 + 3 / 1 6 = 3 / 8