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If Ben were to lose the championship, Mike would be the winner with a

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Joined: 17 Apr 2013
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GMAT 1: 710 Q50 V36
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Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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New post 01 Aug 2015, 01:40
Bunuel wrote:
asmit123 wrote:
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 being the winner is \(\frac{1}{7}\), what is the probability that either Mike or Rob will win the championship? Assume that there can be only one winner.

A. \(\frac{1}{12}\)
B. \(\frac{1}{7}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{12}\)
E. \(\frac{6}{7}\)


M07-12


This is a conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must lose: the probability of Ben losing is \(1-\frac{1}{7}=\frac{6}{7}\).

Now out of these \(\frac{6}{7}\) cases the probability of Mike winning is \(\frac{1}{4}\) and the probability of Rob winning is \(\frac{1}{3}\). So \(P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}\).

Or consider the following:

Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).

Now, out of these 84 cases Ben will lose in \(\frac{6}{7}*84=72\). Mike would be the winner in \(72*\frac{1}{4}=18\) (1/4 th of the cases when Ben loses) and Rob would be the winner in \(72*\frac{1}{3}=24\). Therefore \(P=\frac{18+24}{84}=\frac{1}{2}\).

Answer: C.


Bunuel what is the problem in my understanding -
I solved it like this -

6/7X1/4X2/3 + 6/7X3/4X1/3 = 5/14
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Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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New post 18 Nov 2017, 05:56
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Re: If Ben were to lose the championship, Mike would be the winner with a   [#permalink] 18 Nov 2017, 05:56

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