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A family has 3 children. Assume that each child is as likely to be a boy as it is to be a girl. Find the probability that the family has 3 girls if it is known the family has at least one girl.

Outcomes: GGG, BBB, GGB, GBG, GBB, BBG, BGB, BGG (8 in total)

When I first approached this problem, I simply did (1)(1/2)(1/2) = 1/4 because I thought there was a 100% certainty that one of the children is a girl; hence a probability of 1. And then the remaining children have a 50% chance of being either a girl or boy. 1/4 is the correct answer, but I now do not believe this way of thinking was correct.

My second approach was to view this problem as a conditional probability where P(A|B) = P(A and B)/P(B) where event A is the family having 3 girls and event B is family having at least one girl.

Therefore, P(A|B) = (1/8) / (7/8) where 1/8 is GGG and 7/8 has at least one girl using the complement. But this is obviously incorrect because my online homework does not accept this as an answer. It accepts 1/4.

I saw an online solution with similar question except for boys and it showed the solution using conditional probability as: (1/8) / (1/8 + 3/8) = 1/4. I don't understand how P(B) = 1/8 + 3/8. Here is the link: https://www.quora.com/A-family-has-three-children-What-is-the-probability-that-they-are-all-boys-given-that-at-least-one-of-them-is-a-boy

Any help in plain, simple language is appreciated!

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    It seems the "online solution with similar question" really is not so similar after all. We can't explain how $P(B) = 1/8 + 3/8$ (at least, we can't be _sure_ we're correctly explaining it) if you don't edit your text above to show the exact wording of that question (or at least show a link to the online solution). In the absence of a clear statement for that problem, I would make the same guess as Brian Scott. – 2017-01-11
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    Some of the on-line schools (you did not mention which one is involved here) have pretty poor math problems/answers. The second approach is the only reasonable interpretation of the problem as described. – 2017-01-12

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The answer that you dismiss as ‘obviously incorrect’ is correct. The fact that the family has at least one girl rules out exactly one of the $8$ otherwise equally likely possibilities, namely the BBB possibility. That means that the actual sample space has $7$ equally likely possibilities, only one of which is $3$ girls, so the probability that the family has $3$ girls is $\frac17$.

The online solution that you mention is either wrong or for a problem that differs in more than just changing girls to boys. It would be correct, for example, for the probability that a family with with three kids, at least two of whom are boys, has three boys.

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    My online homework accepted 1/4 as an input but claims that 1/7 is incorrect. Not sure why.. – 2017-01-11
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    @J.K: $\frac14$ would be the correct answer if we knew that a specific one of the children was a girl. For instance, if we know that the eldest child is a girl, then the probability that all three are girls is $\frac14$, since the sample space now has just the four elements BB, BG, GB, and GG (second child, third chile). – 2017-01-11
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    I guess the wording of the problem was a bit ambiguous. Thank you! – 2017-01-11
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    @J.K: You’re welcome! – 2017-01-11