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Xavier, Yvonne, and Zelda each try independently to solve a

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Xavier, Yvonne, and Zelda each try independently to solve a [#permalink] New post 19 Jun 2008, 23:58
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Oct 2012, 02:56, edited 1 time in total.
Renamed the topic, edited the question and moved to PS forum.
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Re: OG C 231 [#permalink] New post 20 Jun 2008, 00:10
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OG C# 231
Xavier, Yvonne, Zelda each try independently to solve a problem. If their individual probabilities
for success are 1/4, 1/2, and 5/8, repectively, what is the probability that xavier and yvonne, but
not zelda, will solve the problem? Please provide solution with explanation. THanks
a. 11/8
b. 7/8
c. 9/64
d. 5/64
e. 3/64


P(Xavier will solve)=1/4
P(Yvonne will solve)=1/2
P(Zelda will NOT solve) = 1- 5/8 = 3/8.


Now, we need to multiply all this Ps to find an answer:
p= (1/4)*(1/2)*(3/8) = 3/64.

Ans. E.
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Re: OG C# 231 Xavier, Yvonne, Zelda each try independently to [#permalink] New post 12 Oct 2012, 08:36
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Yes - since they are independent events, you can multiply the probabilities together.

Be sure to remember to take the opposite probability for that last one.

Please see: http://www.gmatpill.com/gmat-practice-t ... stion/2387


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Re: OG C 231 [#permalink] New post 02 May 2013, 14:46
greenoak wrote:
Quote:
OG C# 231

Now, we need to multiply all this Ps to find an answer:
p= (1/4)*(1/2)*(3/8) = 3/64.


I always get confused, Why don't we multiply it by 3? because these independent event can occur is any order. that means it can also be (3/8)*(1/4)*(1/2) ?
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Re: OG C 231 [#permalink] New post 16 May 2013, 13:53
I'm not great with probability so double check whatever I say, but we're not looking for the order in which X and Y get it right and Z get's it wrong. We're just looking for the probability of X and Y getting it right while Z get's it wrong. It doesn't matter in what order they get it right or wrong, all that matters is how they perform. Think of it like flipping a coin a few times and recording the probability of getting heads. It doesn't matter if you get two heads in a row then one tails, or one tails then two heads in a row. The probability is the same either way. This is in contrast to say, the chances of gettign a certain color gum ball in a certain order.

Hope that helps!

nikhil007 wrote:
greenoak wrote:
Quote:
OG C# 231

Now, we need to multiply all this Ps to find an answer:
p= (1/4)*(1/2)*(3/8) = 3/64.


I always get confused, Why don't we multiply it by 3? because these independent event can occur is any order. that means it can also be (3/8)*(1/4)*(1/2) ?
Re: OG C 231   [#permalink] 16 May 2013, 13:53
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