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Probability that Mike OR Rob wins is 1/4 + 1/3 = 7/12.
Probability that Ben does NOT win = 5/6.
Thus probability that Mike or Rob wins, and Ben does not = 35/72.
_________________

Probability that Mike OR Rob wins is 1/4 + 1/3 = 7/12. Probability that Ben does NOT win = 5/6. Thus probability that Mike or Rob wins, and Ben does not = 35/72.

This question is from a GMAT Challenge test. I got the same answer as you but the computer said it was 5/12. I just want to confirm that we have the right answer.

where A - Probability of A winning and (A) - Probability of A not winning

=> 1/4*1/3*5/6 + 1/4*2/3*5/6 + 3/4*1/3*5/6 = 5/12

- Vipin

Good explanatio Vipin.
I also fell into the trap
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

Hi Vipin, Thanks for your explanation but we need scenario 1 based on the question. Why do we need to consider rest of the scenarios i mean 2and3?

ma466

Hi MA,
Actually question says Mike OR Rob should win and Bob shouldn't win. Had it been Mike AND Rob should win, then we would have considered only Scenario1.

The answers seems correct but what about the approach?

Well, it's interesting that you arrived at the correct solution,
but I fear you were lucky

First event: The probability that Ben doesn't win is
1-1/6 = 5/6

Second event: The probability that Mike or Rob wins is
1/4 + 1/3 - (1/4 * 1/3) = 1/2

So the answer is 5/6 * 1/2 = 5/12.

Those of you who got 35/72 forgot to subtract 1/3 * 1/4 from
the addition Mike and Rob. This has to be done because this
part has been counted twice.

Just consider two events with probability 0.8 that are independent.
The probability that at least one of those occurs is
0.8 + 0.8 - 0.8*0.8 = 0.96 (and of course not 1.6).