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Xavier, Yvonne, and Zelda each try independently to solve a
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03 Sep 2020, 10:26
03 Sep 2020, 10:26
Expert Reply
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Question Stats:
86% (00:55) correct
13% (02:00) wrong based on 37 sessions
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Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2 and 5/8, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?
A. 11/8
B. 7/8
C. 9/64
D. 5/64
E. 3/64
A. 11/8
B. 7/8
C. 9/64
D. 5/64
E. 3/64
Re: Xavier, Yvonne, and Zelda each try independently to solve a
[#permalink]
03 Sep 2020, 10:26
03 Sep 2020, 10:26
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
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Re: Xavier, Yvonne, and Zelda each try independently to solve a
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03 Sep 2020, 11:29
03 Sep 2020, 11:29
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Carcass wrote:
Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2 and 5/8, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?
A. 11/8
B. 7/8
C. 9/64
D. 5/64
E. 3/64
A. 11/8
B. 7/8
C. 9/64
D. 5/64
E. 3/64
P(Z solves the problem) = 1 - P(Z doesn't solve the problem)
So, 5/8 = 1 - P(Z doesn't solve the problem)
So, P(Z doesn't solve the problem) = 3/8
The question asks us to find P(Xavier and Yvonne solve problem, but Zelda does not solve problem)
So, we want: P(X solves problem AND Y solves problem AND Z does not solve)
= P(X solves problem) x P(Y solves problem) x P(Z does not solve)
= 1/4 x 1/2 x 3/8
= 3/64
Answer: E
