How can "there's $\le$ 1 girl in the family" be independent from "the family has children of both genders"?

Solution 1:

Since you have already received two detailed correct answers I will try to address what I think might be confusing about the conditions in this question.

You are given two events, one of them $B$ is gender-indifferent (it describes an event which says the same thing about both genders), whereas the other one, $A$, breaks the gender symmetry, and it seems intuitively impossible for two such events to be independent.

However, in the specific conditions of the question it is worth noting that the complement of $A$ is the event in which the family has at least two daughters, which is the same as saying it has at most one son, and is therefore the same as $A$ after switching the genders. So whether $A$ occurs or not is equivalent to choosing between two gender-symmetrical alternatives, and hence cannot change the probability of another, gender-symmetrical event, such as $B$.

Solution 2:

The two events are independent.

If two events $A$ and $B$ are independent, then by definition, $P(A\cap B)=P(A)\cdot P(B)$.

The probability of $A$ (the family having at most one girl) is $1/2$, since there is one way to have no girls and three ways to have one girl in the family, out of the eight possible cases.

The probability of $B$ (the family has both genders) is $3/4$, since there are only two ways to have only one gender in the family, out of the eight cases.

The probability of $A\cap B$ (the family has at most one girl and has both genders) is $3/8$, since the three cases with one girl ensures the family has both genders.

Checking $P(A\cap B)=P(A)\cdot P(B)$, we have $3/8\overset{?}{=}(1/2)(3/4)$, which is true. Hence, by the definition of independent events, $A$ and $B$ are independent.