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probability :wrestling

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probability :wrestling

by psm12se » Wed Jan 01, 2014 1:20 am
John and Tina wanted to place a bet for fun on one of three wrestling matches last night. Last night, Wrestler A defeated Wrestler B. Wrestler B defeated Wrestler C, and Wrestler C defeated Wrestler A. Assuming that both John and Tina had an equal chance of choosing any one of the three wrestlers to bet on, what is the probability that John won the bet?

A. 5/6
B. 2/3
C. 1/2
D. 5/12
E. 1/3

Got puzzled on this one. Please anyone tell how to approach this problem.
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by Brent@GMATPrepNow » Wed Jan 01, 2014 8:13 am
psm12se wrote:John and Tina wanted to place a bet for fun on one of three wrestling matches last night. Last night, Wrestler A defeated Wrestler B. Wrestler B defeated Wrestler C, and Wrestler C defeated Wrestler A. Assuming that both John and Tina had an equal chance of choosing any one of the three wrestlers to bet on, what is the probability that John won the bet?

A. 5/6
B. 2/3
C. 1/2
D. 5/12
E. 1/3

Got puzzled on this one. Please anyone tell how to approach this problem.
I'm not entirely sure what this question is asking.
Both John and Tina had an equal chance of choosing any one of the three wrestlers to bet on
Does this mean that they could both bet on Wrestler A in the first match? In this case, neither John nor Tina would with the bet.
Even weirder, it seems that they could both bet on Wrestler C in the first match (even though this wrestler is not in the first match).
Even weirder...er, it seems that John could both bet on Wrestler A in the first match, and Tina could both bet on Wrestler B in the second match in which case John and Tina would both win.

If John and Tina must select different wrestlers in the same match, then the probability is 1/2 that John won the bet (since there are 2 wrestlers and only 1 of them wins).

Given the ambiguity of this question, it's safe to say that it would never appear in an official GMAT.

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Wed Jan 01, 2014 9:37 am, edited 1 time in total.
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by GMATGuruNY » Wed Jan 01, 2014 9:07 am
I agree with Brent: it's not clear what the problem above is asking.
A similar -- but much more clearly worded -- problem:

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