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!