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Two canoe riders must be selected from each of two groups of

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Two canoe riders must be selected from each of two groups of [#permalink] New post 11 Aug 2003, 05:27
Two canoe riders must be selected from each of two groups of campers.
One group consists of three men and one woman, and the other group consists of two women and one man. What is the probability that two men and two women will be selected?
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Re: Probability [#permalink] New post 11 Aug 2003, 06:34
mistdew wrote:
Two canoe riders must be selected from each of two groups of campers.
One group consists of three men and one woman, and the other group consists of two women and one man. What is the probability that two men and two women will be selected?


P1 (two men out of 3 men and 1 woman) = 3/4*2/3=1/2
P2 (two women out of 2 women and 1 men) = 2/3*1/2=1/3

P3 (P1&P2) =1/2*1/3=1/6
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 [#permalink] New post 11 Aug 2003, 14:46
I disagree. There are these two:

P1 (two men out of 3 men and 1 woman) = 3/4*2/3=1/2
P2 (two women out of 2 women and 1 men) = 2/3*1/2=1/3

but also P3 (one each from both canoes) = 1/2*2/3=1/3.

So IMO, the answer is 1/2*1/3 + 1/3 = 1/2.
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 [#permalink] New post 11 Aug 2003, 17:39
Akami...you are absolutely right....but pls could you(or anyone) explain the P3 part?

"but also P3 (one each from both canoes) = 1/2*2/3=1/3. "


Thanks
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 [#permalink] New post 11 Aug 2003, 21:00
mistdew wrote:
Akami...you are absolutely right....but pls could you(or anyone) explain the P3 part?

"but also P3 (one each from both canoes) = 1/2*2/3=1/3. "


Thanks


Canoe 1: 4c2 ways o r6 to pull two people. 3 of those ways were 2 men. so 3/6 to pull one man and one women.

Canoe 2: 3c2 ways or 3 ways to pull two people. One of those ways was two women. Hence 2/3 to pull one each.

1/2 * 1/3 = 1/6
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Former Senior Instructor, Manhattan GMAT and VeritasPrep
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MBA, Anderson School of Management, UCLA, Class of 1993

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 [#permalink] New post 21 Jan 2004, 07:49
sligtly different approach for a very old problem

Group 1 : 3M 1W
Group 2 : 1M 2W

Select two from each such that 2M and 2W
From Group1 2M can be selected in 3C2
From Group2 2W can be selected in 2C2
So total = 3C2*2C2 = 3
From group 1 select 1M 1W in 3C1*1C1
From group 2 select 1M 1W in 1C1*2C1
Total = 3C1*1C1 * 1C1*2C1 = 6

Total ways to select 2 from each group = 4C2 * 3C2 = 18

Probability = (3+6)/18 = 1/2
  [#permalink] 21 Jan 2004, 07:49
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