Win just once

The probability that Marvin will win in a certain game whenever he plays is 1 4 \dfrac14 . If he plays twice, find the probability that he will win just once.

3 8 \frac38 1 8 \frac18 3 18 \frac3{18} 3 16 \frac3{16}

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1 solution

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 / 16 1/4 * 3/4 = 3/16

P(b) = 3 / 4 1 / 4 = 3 / 16 3/4 * 1/4 = 3/16

P(a) or P(b) = 3 / 16 + 3 / 16 = 3 / 8 3/16 + 3/16 = 3/8

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