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The probability that Mike can win is 1/4, Rob winning is

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The probability that Mike can win is 1/4, Rob winning is [#permalink] New post 18 Apr 2006, 11:58
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
1
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Re: Probability [#permalink] New post 18 Apr 2006, 12:19
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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Re: Probability [#permalink] New post 18 Apr 2006, 12:21
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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 [#permalink] New post 18 Apr 2006, 13:14
yeah I can't reach any other answer
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Re: Probability [#permalink] New post 19 Apr 2006, 05:50
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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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 [#permalink] New post 19 Apr 2006, 06:12
I also agree with Kpslock anser explanation
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 [#permalink] New post 19 Apr 2006, 06:34
Vipin,
Could you please explain your answer? I did not unsderstand your answer. Thanks
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 [#permalink] New post 19 Apr 2006, 07:13
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,
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Re: Probability [#permalink] New post 19 Apr 2006, 07:26
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 :wall :wall

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 [#permalink] New post 19 Apr 2006, 07:31
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?

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 [#permalink] New post 19 Apr 2006, 08:39
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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 [#permalink] New post 20 Apr 2006, 10:20
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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 [#permalink] New post 22 Apr 2006, 05:11
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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 :-D

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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 [#permalink] New post 23 Apr 2006, 09:33
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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 [#permalink] New post 24 Apr 2006, 04:27
Great explanation ccax. :)
  [#permalink] 24 Apr 2006, 04:27
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