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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]

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18 Apr 2006, 11:58
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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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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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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18 Apr 2006, 13:14
yeah I can't reach any other answer
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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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19 Apr 2006, 06:12
I also agree with Kpslock anser explanation
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19 Apr 2006, 06:34
Vipin,
ma466
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19 Apr 2006, 07:13
MA466 wrote:
Vipin,
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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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
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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?

ma466
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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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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

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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

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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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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24 Apr 2006, 04:27
Great explanation ccax.
24 Apr 2006, 04:27
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