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set 37 - #16

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set 37 - #16

by KL08 » Sun Sep 23, 2007 8:34 pm
16. 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

OA is E
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by samirpandeyit62 » Sun Sep 23, 2007 9:21 pm
Xavier and Yvonne, but not Zelda solve the problem

=P(Xavier solves)*P( Yvonne) *P (Zelda doesn't solve)

=1/4 *1/2 * 3/8

=3/64
Regards
Samir
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by uptowngirl92 » Sat Oct 03, 2009 12:03 am
I got a querie here:
Why are we not doing this?
X and Y solve the problem = 1-(Z does not solve the problem).
=1-3/8=5/8 :!:
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by sanjana » Mon Oct 05, 2009 10:10 am
Uptown girl,
Z not solving the problem doesnt mean X and Y have solved the problem. It is no where mentioned that X AND Y solving the problem is dependent on Z solving the problem or not.
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