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

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Director
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 532
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Re: If Ben were to lose the championship, Mike would be the winner with a  [#permalink]

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01 Aug 2015, 02:40
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 $$P=\frac{18+24}{84}=\frac{1}{2}$$.

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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18 Nov 2017, 06:56
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Re: If Ben were to lose the championship, Mike would be the winner with a &nbs [#permalink] 18 Nov 2017, 06:56

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