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

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11 Jul 2003, 19:38

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

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

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