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The probability that Mike can win is 1/4, Rob winning is [#permalink]
 18 Apr 2006, 12:58
The probability that Mike can win is 1/4, Rob winning is 1/3, Ben winning is 1/6. What is the probability that mike or rob will win but NOT Ben.
1/6
5/12
7/12
5/6
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Senior Manager
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Am I making a mistake?
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.
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Who says elephants can't dance?
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Intern
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kapslock wrote: Am I making a mistake?
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.
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Senior Manager
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yeah I can't reach any other answer
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yach wrote: The probability that Mike can win is 1/4, Rob winning is 1/3, Ben winning is 1/6. What is the probability that mike or rob will win but NOT Ben.
1/6
5/12
7/12
5/6
1
There are 3 scenarios
MR(B)
M(R)(B)
(M)R(B)
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
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I also agree with Kpslock anser explanation
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Manager
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Vipin,
Could you please explain your answer? I did not unsderstand your answer. Thanks
ma466
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MA466 wrote: Vipin,
Could you please explain your answer? I did not unsderstand your answer. Thanks
ma466
There are 3 scenarios. Either Mike and Rob will win, or only Mike will win, or only Rob will win.
p(M wins) = 1/4 => p(M doesn't win) = 3/4
p(R wins) = 1/3 => p(R doesn't win) = 2/3
p(B wins) = 1/6 => p(B doesn't win) = 5/6
Scenario1 -
p(M wins) * p(R wins) * p(B doesn't win) +
Scenario2
p(M wins) * p(R doesn't win) * p(B doesn't win) +
Scenario3
p(M doesn't win) * p(R wins) * p(B doesn't win)
This is what the above expression represents.
HTH...
Thanks,
Vipin
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VP
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vipin7um wrote:
There are 3 scenarios
MR(B)
M(R)(B)
(M)R(B)
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 
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"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."
- Bernard Edmonds
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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
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Senior Manager
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MA466 wrote: 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.
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Hi! I am newbie - both to gmat and this forum.. i arrived at the solution like this:
(1/4 + 1/3) - 1/6
which is 5/12
The answers seems correct but what about the approach?
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rambler wrote: i arrived at the solution like this:
(1/4 + 1/3) - 1/6
which is 5/12
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).
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Intern
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I have nothing more to add to ccax post. I also arrive to the solution in the same way.
(P(M) + P(R) - P(M)(R)) * P(not B) = 5/12
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Great explanation ccax. 
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