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# M07#12

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Intern
Joined: 06 May 2008
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17 Jul 2009, 08:15
Why the answer doesn't make much sense to me??

I strongly believe that since the probability of Mike or Rob winning, conditional on Ben losing, is 7/12, then that means that:
7/12=(1-1/7)*(x+y), where x and y are the probabilities of Mike or Rob winning respectivelly. That would mean x+y=49/72.

The OE says that x+y=(1-1/7)*7/12...
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20 Jul 2009, 00:08
DFG5150 wrote:
Why the answer doesn't make much sense to me??

I strongly believe that since the probability of Mike or Rob winning, conditional on Ben losing, is 7/12, then that means that:
7/12=(1-1/7)*(x+y), where x and y are the probabilities of Mike or Rob winning respectivelly. That would mean x+y=49/72.

The OE says that x+y=(1-1/7)*7/12...

I am not sure I get what you mean. here is a copy of the OE just in case - which part do you disagree with?

The probability of Mike or Rob winning, conditional on Ben losing, is $$\frac{1}{4} + \frac{1}{3}$$ or $$\frac{7}{12}$$ . The unconditional probability (to take into consideration the probability of Ben winning) is then $$(1-\frac{1}{7}) * \frac{7}{12} = \frac{6}{7} * \frac{7}{12} = \frac{1}{2}$$ .
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Senior Manager
Joined: 08 Jun 2010
Posts: 397
Location: United States
Concentration: General Management, Finance
GMAT 1: 680 Q50 V32
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25 Apr 2012, 23:47
quick question on this.

There are two scenarios where either Mike or Rob can win.

B(L) x M(W) x R(L)
B(L) x M(L) x R(W)

We know both can't win.

= (6/7)*(1/4)*(2/3) + (6/7)*(3/4)*(1/3)
= 5/14.

Why do we not include Rob's result in the answer? We are constrained by one winner so Rob's win or loss will affect the probability, correct?
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26 Apr 2012, 00:55
mourinhogmat1 wrote:
quick question on this.

There are two scenarios where either Mike or Rob can win.

B(L) x M(W) x R(L)
B(L) x M(L) x R(W)

We know both can't win.

= (6/7)*(1/4)*(2/3) + (6/7)*(3/4)*(1/3)
= 5/14.

Why do we not include Rob's result in the answer? We are constrained by one winner so Rob's win or loss will affect the probability, correct?

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. $$\frac{6}{7}$$

This is conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must loose --> the probability of Ben loosing 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 loose 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 loose) and Rob would be the winner in $$72*\frac{1}{3}=24$$ --> $$P=\frac{18+24}{84}=\frac{1}{2}$$.

In case of any question please post it here: if-ben-were-to-lose-the-championship-m07q12-75695.html
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Re: M07#12   [#permalink] 26 Apr 2012, 00:55
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# M07#12

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