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

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Current Student
Joined: 31 Aug 2007
Posts: 370
Followers: 1

Kudos [?]: 49 [0], given: 1

Two canoe riders must be selected from each of two groups of [#permalink]  30 Sep 2007, 15:10
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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
Director
Joined: 03 May 2007
Posts: 894
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 61 [0], given: 7

Re: PS probability [#permalink]  30 Sep 2007, 15:51
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...
Manager
Joined: 10 Jan 2005
Posts: 63
Followers: 1

Kudos [?]: 4 [0], given: 0

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
Director
Joined: 22 Aug 2007
Posts: 572
Followers: 1

Kudos [?]: 14 [0], given: 0

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
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5079
Location: Singapore
Followers: 21

Kudos [?]: 180 [0], given: 0

# 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
Manager
Joined: 10 Jan 2005
Posts: 63
Followers: 1

Kudos [?]: 4 [0], given: 0

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.
Director
Joined: 22 Aug 2007
Posts: 572
Followers: 1

Kudos [?]: 14 [0], given: 0

Re: PS probability [#permalink]  30 Sep 2007, 22:32
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..
Current Student
Joined: 31 Aug 2007
Posts: 370
Followers: 1

Kudos [?]: 49 [0], given: 1

ywilfred is correct, OA is 1/2.
VP
Joined: 08 Jun 2005
Posts: 1147
Followers: 6

Kudos [?]: 127 [0], given: 0

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 !

Director
Joined: 03 May 2007
Posts: 894
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 61 [0], given: 7

Re: PS probability [#permalink]  01 Oct 2007, 06:36
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
Director
Joined: 22 Aug 2007
Posts: 572
Followers: 1

Kudos [?]: 14 [0], given: 0

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?
Manager
Joined: 01 Oct 2007
Posts: 87
Followers: 0

Kudos [?]: 14 [0], given: 0

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.
Manager
Joined: 10 Jan 2005
Posts: 63
Followers: 1

Kudos [?]: 4 [0], given: 0

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.
Director
Joined: 18 Jul 2006
Posts: 531
Followers: 1

Kudos [?]: 26 [0], given: 0

Re: PS probability [#permalink]  01 Oct 2007, 14:34
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, 14:34
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