Choice Method vs Probability Fractions for Probability x Combinatorics

[user4341]

“A couple has five young children. Given that at least one of the children is a girl, what is the probability that exactly three of the five children are girls?”

The given solution was to use the choice method for the total permutations, assuming 2^5 (_ _ _ _ _ with 2 choices - Boy or girl- for each) = 32 total outcomes, and then to subtract the BBBBB case = 31 → this becomes the denominator for the probability.

When are we able to use this approach vs having to do the (1/2)^5 for each blank - is that just when the probabilities are equal like for Boy/Girl?

[cylverixxx]

I mean the complement of “at least one girl” is “no girls”, which is why we do 1 - \frac{1}{32}, where the sole arrangement (among all 32 possible arrangements) that doesn’t involve a girl is BBBBB.

The rest is just applying the above formula for conditional probability.

In particular, P(B) = \frac{31}{32} and P(A \cap B) = \frac{10}{32} thus giving you P(A | B) = \frac{10}{31}