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Q2

by Suyog » Mon Sep 24, 2007 2:09 pm
Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are , , and , 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
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by Suyog » Mon Sep 24, 2007 2:43 pm
i'm sorry....
didnt realize....

Xavier 1/4
Yvonne 1/2
and Zelda 5/8
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by niks_01.27 » Mon Sep 24, 2007 3:52 pm
Probability of Xavier (X) solving the problem = 1/4
Probability of Yvonne (Y) solving the problem = 1/2

Probability of both X & Y solving the problem = 1/4 x 1/2 = 1/8

Probability of both Zelda (Z) not solving the problem = 1 - 5/8 = 3/8

Now, Probability of both X & Y solving the problem and Z not solving the problem = 1/8 x 3/8 = 3/64

Answer should be E.
regards
niks...
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by Suyog » Mon Sep 24, 2007 4:02 pm
Thanks niks!
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