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# Ms. Barton has four children. You are told correctly that

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GMAT Instructor
Joined: 07 Jul 2003
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Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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Ms. Barton has four children. You are told correctly that [#permalink]  11 Jul 2003, 18:38
Ms. Barton has four children. You are told correctly that she has a least two girls but you are not told which two of her four children are those girls. What is the probability that she also has two boys?
(A) 1/4
(B) 3/8
(C) 5/11
(D) 1/2
(E) 6/11
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager
Joined: 18 Jun 2003
Posts: 141
Location: Hockeytown
Followers: 2

Kudos [?]: 16 [0], given: 0

Ms. Barton has at least two girls, but if she is to also have two boys, she must have two girls:

BBGG

Assuming the probability of having a girl and a boy is equal (which is still a fair assumption, I believe) and = 1/2,

P(2 boys and 2 girls) = (1/2)^4 x 4c2 = 1/16 x 6 = 6/16 = 3/8

If the question were to call for the P(at least 1 boy):

P = 3/8 + [1/16 x 4c1] = 3/8 + 1/4 = 5/8
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

Kudos [?]: 52 [0], given: 0

JP wrote:
Ms. Barton has at least two girls, but if she is to also have two boys, she must have two girls:

BBGG

Assuming the probability of having a girl and a boy is equal (which is still a fair assumption, I believe) and = 1/2,

P(2 boys and 2 girls) = (1/2)^4 x 4c2 = 1/16 x 6 = 6/16 = 3/8

If the question were to call for the P(at least 1 boy):

P = 3/8 + [1/16 x 4c1] = 3/8 + 1/4 = 5/8

Sorry, that is NOT correct. Your assumption that having a boy and girl are equally probable is reasonable. Your approach is mostly correct but you are forgetting something....
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager
Joined: 18 Jun 2003
Posts: 141
Location: Hockeytown
Followers: 2

Kudos [?]: 16 [0], given: 0

I think I may have misinterpreted the question. We can look at it like this:

4 children
at least 2 daughers => 2, 3, or 4 daughters

2 daughters => 2 boys => 4c2 combinations = 6 combinations
3 daughters => 1 boy => 4c1 combinations = 4 combinations
4 daughters => 0 boys => 4c0 combinations = 1 combination

So there are a total of 11 combinations, of which 6 leave room for 2 boys. Thus 6/11.
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

Kudos [?]: 52 [0], given: 0

JP wrote:
I think I may have misinterpreted the question. We can look at it like this:

4 children
at least 2 daughers => 2, 3, or 4 daughters

2 daughters => 2 boys => 4c2 combinations = 6 combinations
3 daughters => 1 boy => 4c1 combinations = 4 combinations
4 daughters => 0 boys => 4c0 combinations = 1 combination

So there are a total of 11 combinations, of which 6 leave room for 2 boys. Thus 6/11.

Okay, you got it this time. Since this is a "conditional" probability problem, the "universe" must be narrowed down first.

Good job!
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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