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Xavier, Yvonne, and Zelda each try independently to solve a [#permalink]
 20 Jun 2008, 00:58
Question Stats:
86% (01:32) correct
13% (01:23) wrong based on 28 sessions
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
Last edited by Bunuel on 13 Oct 2012, 03:56, edited 1 time in total.
Renamed the topic, edited the question and moved to PS forum.
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Director
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Quote: 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]
12 Oct 2012, 09:36
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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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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) ? |
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