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# m07 q12

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m07 q12 [#permalink]  02 Oct 2010, 05:57
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0% (00:00) correct 100% (00:43) wrong based on 3 sessions
If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$ , and Rob - $$\frac{1}{3}$$ . If the probability of Ben winning is $$\frac{1}{7}$$ , what is the probability that either Mike or Rob will win the championship (there can be only one winner)?

(C) 2008 GMAT Club - m07#12

* $$\frac{1}{12}$$
* $$\frac{1}{7}$$
* $$\frac{1}{2}$$
* $$\frac{7}{12}$$
* $$\frac{6}{7}$$
[Reveal] Spoiler: OA
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Intern
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Re: m07 q12 [#permalink]  02 Oct 2010, 06:00
The OA : $$(1-\frac{1}{7}) * \frac{7}{12} = \frac{6}{7} * \frac{7}{12} = \frac{1}{2}$$ .
(7/12 = 1/3 + 1/4)

Can Some one explain me what is the problem with this reasoning :
P=(1/4 *2/3 * 6/7) + (1/3 * 3/4 * 6/7)= 5/14

Thanks
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Re: m07 q12 [#permalink]  02 Oct 2010, 07:12
Barkatis wrote:
The OA : $$(1-\frac{1}{7}) * \frac{7}{12} = \frac{6}{7} * \frac{7}{12} = \frac{1}{2}$$ .
(7/12 = 1/3 + 1/4)

Can Some one explain me what is the problem with this reasoning :
P=(1/4 *2/3 * 6/7) + (1/3 * 3/4 * 6/7)= 5/14

Thanks

I think what you have done is :

Probability = P(Ben loses) * P(Mike wins given Ben loses) * (1 - P(Rob wins givne Ben loses)) + P(Ben loses) * P(Rob wins given Ben loses) * (1 - P(Mike wins givne Ben loses))

The problem is the terms I have marked in red. You are already given the probability that Mike wins once Ben has lost, 1/4, you do not need to multiply this with (1-2/3). Similarly in the second term

The correct answer would be : P(Ben loses) * P(Mike wins given Ben loses) + P(Ben loses) * P(Rob wins given Ben loses)
OR (6/7) * (1/3) + (6/7) * (1/4) = (1/2)
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Re: m07 q12 [#permalink]  02 Oct 2010, 12:11

I got that part but my question is how do we know that the probability given is that Mike wins ONCE BEN HAS LOST.

What is the difference between that problem and this one for example
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 ?

Where the solution is 3/8*1/2*1/4 = 3/64
But if we assume that the probability given are those of success of one person while the two others are loosing it would be more : 1/4*1/2 . Right ?

I hope you got my problem better now.
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Re: m07 q12 [#permalink]  02 Oct 2010, 17:26
Barkatis wrote:

I got that part but my question is how do we know that the probability given is that Mike wins ONCE BEN HAS LOST.

What is the difference between that problem and this one for example
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 ?

Where the solution is 3/8*1/2*1/4 = 3/64
But if we assume that the probability given are those of success of one person while the two others are loosing it would be more : 1/4*1/2 . Right ?

I hope you got my problem better now.

You have to be careful about reading the wording. In the example you are giving, there are 3 people solving a question. None, one or more could solve it correctly. In the question at hand, there are 3 players trying to win a tournament. Either none of them or atmost one of them can win, and simultaneous winning is not possible.

Now whether we have the probability of Mike winning given Ben has lost or we have absolute probability of Mike winning is just how the question is worded.

If Ben were to lose the championship, Mike would be the winner with a probability of \frac{1}{4}

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Re: m07 q12   [#permalink] 02 Oct 2010, 17:26
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# m07 q12

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