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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]

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30 Sep 2007, 16:10
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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?

(A) 1/6
(B) 1/4
(C) 2/7
(D) 1/3
(E) 1/2
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30 Sep 2007, 16:51
young_gun 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?

(A) 1/6
(B) 1/4
(C) 2/7
(D) 1/3
(E) 1/2

hmmmmmmmm........ i got something else.

total = 4c2 x 3c2 = 12
2m2w = 3c2x2c2 = 3
1m1w from 1st and 1m1w from 2nd = (3x1) x (2x1) = 6
so total = 9
the prob = 9/12 = 3/4 which is not the answer...
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30 Sep 2007, 17:39
I'm going with A.

There are only two possibilities:
1) Choose 2 men from group 1 and 2 women from group 2
2) Choose 1 man, 1 woman from group 1 and 1 man, 1 woman from group 2

So the total probability is:

P(2m, 2w) OR P(1m1w, 1m1w)

P(2m, 2w) = (3/4)*(1/3) AND (2/3)*(1/2) = 1/12

P(1w1m, 1w1m) = (1/4)(1) AND (1/3)*(1) = 1/12

add these because of the OR: 2/12 = 1/6
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30 Sep 2007, 18:39
mayonnai5e wrote:
I'm going with A.

There are only two possibilities:
1) Choose 2 men from group 1 and 2 women from group 2
2) Choose 1 man, 1 woman from group 1 and 1 man, 1 woman from group 2

So the total probability is:

P(2m, 2w) OR P(1m1w, 1m1w)

P(2m, 2w) = (3/4)*(1/3) AND (2/3)*(1/2) = 1/12

P(1w1m, 1w1m) = (1/4)(1) AND (1/3)*(1) = 1/12

add these because of the OR: 2/12 = 1/6

why is it 1/3 not 2/3 ? After 1 man chosen 2 men are left, in group of 3...?

Agree with this answer otherwise, though i get 3/4, not in AC
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30 Sep 2007, 23:20
# of ways to pick 4 people = 4C2 * 3C2 = 18

Winning cases:
(1 men, 1 woman from group A) and (1 men and 1 woman from group B) = 3C1 * 2C1 = 6
(2 men from group A) and (2 women from group B) = 3C2 = 3

Probability = 9/18 = 1/2
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30 Sep 2007, 23:22
IrinaOK wrote:
mayonnai5e wrote:
I'm going with A.

There are only two possibilities:
1) Choose 2 men from group 1 and 2 women from group 2
2) Choose 1 man, 1 woman from group 1 and 1 man, 1 woman from group 2

So the total probability is:

P(2m, 2w) OR P(1m1w, 1m1w)

P(2m, 2w) = (3/4)*(1/3) AND (2/3)*(1/2) = 1/12

P(1w1m, 1w1m) = (1/4)(1) AND (1/3)*(1) = 1/12

add these because of the OR: 2/12 = 1/6

why is it 1/3 not 2/3 ? After 1 man chosen 2 men are left, in group of 3...?

Agree with this answer otherwise, though i get 3/4, not in AC

I think because of a careless mistake. =)

That 1/3 should be 2/3 to represent the fact that 2 men are left out of the 3 possible choices in group 1 after the first man is selected.

P(2m, 2w) = (3/4)*(2/3) AND (2/3)*(1/2) = 1/6

P(1w1m, 1w1m) = (1/4)(1) AND (1/3)*(1) = 1/12

add these because of the OR: 2/12 + 1/12 = 3/12 = 1/4[/quote]

so perhaps the answer is B.
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30 Sep 2007, 23:32
young_gun 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?

(A) 1/6
(B) 1/4
(C) 2/7
(D) 1/3
(E) 1/2

Such a variety of answers...wonder what OA is..
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01 Oct 2007, 06:27
ywilfred is correct, OA is 1/2.
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01 Oct 2007, 07:03
ywilfred wrote:
# of ways to pick 4 people = 4C2 * 3C2 = 18

Winning cases:
(1 men, 1 woman from group A) and (1 men and 1 woman from group B) = 3C1 * 2C1 = 6
(2 men from group A) and (2 women from group B) = 3C2 = 3

Probability = 9/18 = 1/2

Perfect !

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01 Oct 2007, 07:36
Fistail wrote:
young_gun 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?

(A) 1/6
(B) 1/4
(C) 2/7
(D) 1/3
(E) 1/2

hmmmmmmmm........ i got something else.

total = 4c2 x 3c2 = 12
2m2w = 3c2x2c2 = 3
1m1w from 1st and 1m1w from 2nd = (3x1) x (2x1) = 6
so total = 9
the prob = 9/12 = 3/4 which is not the answer...

Thats what I was struggling:

total = 4c2 x 3c2 = 18
2m2w = 3c2x2c2 = 3
1m1w from 1st and 1m1w from 2nd = (3x1) x (2x1) = 6
so total = 9
the prob = 9/18 = 1/2
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01 Oct 2007, 10:15
mayonnai5e wrote:
I'm going with A.

There are only two possibilities:
1) Choose 2 men from group 1 and 2 women from group 2
2) Choose 1 man, 1 woman from group 1 and 1 man, 1 woman from group 2

So the total probability is:

P(2m, 2w) OR P(1m1w, 1m1w)

P(2m, 2w) = (3/4)*(2/3) AND (2/3)*(1/2) = 1/6

P(1w1m, 1w1m) = (1/4)(1) AND (1/3)*(1) = 1/12

add these because of the OR: 2/12 + 1/12 = 3/12 = 1/4

Can someone explain, plz, why this approach is wrong?
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01 Oct 2007, 12:58
Quote:
mayonnai5e wrote:
I'm going with A.

There are only two possibilities:
1) Choose 2 men from group 1 and 2 women from group 2
2) Choose 1 man, 1 woman from group 1 and 1 man, 1 woman from group 2

So the total probability is:

P(2m, 2w) OR P(1m1w, 1m1w)

P(2m, 2w) = (3/4)*(2/3) AND (2/3)*(1/2) = 1/6

P(1w1m, 1w1m) = (1/4)(1) AND (1/3)*(1) = 1/12

add these because of the OR: 2/12 + 1/12 = 3/12 = 1/4

Can someone explain, plz, why this approach is wrong?

This approach can work, but you have to be careful about defining the probabilities. The definition here for the probability of picking one woman and one man from each group assumes that the woman is picked first. But that's not necessarily true. What we really need is:

P[(1w1m OR 1m1w, 1w1m OR 1m1w)] = [(1/4)*1+ (3/4)(1/3)] * [(1/3)*1+(2/3)(1/2)] = [(1/4) + (1/4)] * [(1/3) +1/3)] = (1/2) * (2/3) = 1/3.

When you add this to the 1/6 from the 2-man situation, you get the correct answer. In this case, it's not the most efficient way to solve the problem (ywilfred's method is more straightforward), but it does get the job done.
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01 Oct 2007, 14:42
johnrb wrote:
P[(1w1m OR 1m1w, 1w1m OR 1m1w)] = [(1/4)*1+ (3/4)(1/3)] * [(1/3)*1+(2/3)(1/2)] = [(1/4) + (1/4)] * [(1/3) +1/3)] = (1/2) * (2/3) = 1/3.

That's just depressing. Imagine doing that on the real CAT.
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01 Oct 2007, 15:34
young_gun 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?

(A) 1/6
(B) 1/4
(C) 2/7
(D) 1/3
(E) 1/2

Agree E.
Earlier I took 2 men and 1 woman in second group....need to be more careful
Re: PS probability   [#permalink] 01 Oct 2007, 15:34
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