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The probability of winning game A is X and the probability of winning
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18 Oct 2021, 05:26
18 Oct 2021, 05:26
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The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once?
A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y
A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y
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Re: The probability of winning game A is X and the probability of winning
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18 Oct 2021, 05:58
18 Oct 2021, 05:58
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Carcass wrote:
The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once?
A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y
A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y
Key concept: P(Event A happening) = 1 - P(Event A not happening)
If P(win game A) = X, then P(lose game A) = 1 - X
If P(win game B) = Y, then P(lose game B) = 1 - Y
P(win exactly once) = P(win game A AND lose game B OR lose game A AND win game B)
= P(win game A) x P(lose game B) + P(lose game A) x P(win game B)
= X x (1 - Y) + (1 - X) x Y
= X - XY + Y - XY
= X + Y - 2XY
Answer: D
Re: The probability of winning game A is X and the probability of winning
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06 Nov 2022, 22:41
06 Nov 2022, 22:41
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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