# Probability that girl who answers the door is the eldest girl?

You know that a family has 3 children. You walk up to and knock on the front door of their house. A girl answers the door.

What is the probability that the she is the eldest girl among the children?

Assume that all 3 children are home and equally likely to answer the door.

(I appreciate that with questions like this, the wording can be so crucial. If the question is in any way ambiguous, I would be delighted to have an explanation of why that is the case!)

• Are you assuming that the probability of having a girl is $50\%$?
– AM13
Commented Dec 2, 2019 at 22:21
• Welcome to Math.SE! What are your own thoughts on this? Commented Dec 2, 2019 at 22:23
• Welcome to stackexchange. You are right to note that the wording is crucial. The best way to get help here (not just an answer) is to edit the question to show us your thoughts. Commented Dec 2, 2019 at 22:23
• To prevent incorrect answers of $\frac{1}{3}$ from being posted again, the answer is clearly greater than $\frac{1}{3}$. This is obvious from the fact that the probability that the girl who answered the door was the eldest child regardless of gender is $\frac{1}{3}$. The girl who answered the door being the eldest girl would also include cases where she were the middle child with an older brother or the youngest child with two older brothers, thus increasing the probability beyond just $\frac{1}{3}$. Commented Dec 3, 2019 at 2:10

There are several correct answers already; I just wanted to give what I think is a simpler argument.

The probability clearly wouldn't change if we instead had a boy answering the door, and wanted to know if he was the oldest boy. So we can equivalently ask what is the probability that the child who answered the door was the oldest child of their sex.

Now there are two possibilities. If all the children are the same sex (probability $$1/4$$), only one child is the oldest of their sex. Otherwise, two children will be the oldest of their sex. Thus the probability is $$\frac14\times\frac13+\frac34\times\frac23=\frac7{12}$$.

This easily generalises to $$n$$ children and gives an answer of $$\frac{2^n-1}{n2^{n-1}}$$.

There are $$8*3=24$$ equally likely situations, the $$8$$ gender orderings and for each one, one of {youngest, middle, eldest} child who answers the door. $$12$$ of those have a girl answering the door (count the $$12$$ G's in the list given in the answer above).

Among those $$12$$ equally likely G's, exactly $$7$$ are the eldest girl (one from each of the $$7$$ non-BBB orderings). So the desired probability is $$7/12$$.

The condition "a random child turns out to be a girl" is NOT equivalent to a YES answer to the question "is there at least one girl?"

• You are mistaken when you state: "Among those 12 equally likely G's." The children were born BEFORE anyone comes to the door. That sets the probability of the family "cases." Next someone comes to the door... The fact that a girl comes to the door does not change the probability that the family HAD HAD three girls or HAD HAD one girl... Commented Dec 3, 2019 at 2:47
• Sorry David, you are wrong about that. Conditional probability doesn't work that way. When a girl answers the door, the (conditional) probability that the family has three girls is more than 1/7 -- in fact it's 1/4 ( = 3/12) since the unknown children at that point are equivalent to flipping two coins. Not the same as being given "at least one girl." It's the same difference between the standard Monte Hall problem vs. the version where a random door is opened.
– Ned
Commented Dec 3, 2019 at 2:56
• Thank you very much for this explanation Ned, it helped me understand the intuition behind the answer. Commented Dec 3, 2019 at 8:22
• @robert_knight Glad to help. Just to be complete, since the "list given above" is no longer here, I meant: BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG, in which 7 of the 12 equally likely G's are eldest.
– Ned
Commented Dec 3, 2019 at 14:22

We seek: the conditional probability, given that a girl in the family answers the door, that this is the eldest.

This will be dependent on how many girls are likely to be in the family, so we shall use the Law of Total Probability and Bayes' Rule.   It is an imperative.

Denote the events as:

• $$G_0,G_1,G_2,G_3$$, that there are 0, 1, 2, or 3 girls in the family, respectively.
• $$A$$, that a girl answers the door (NB: the events $$A,G_0$$ are disjoint)
• $$E$$, that the eldest girl answers the door. (NB: $$E\subseteq A$$, so $$E=E\cap A$$)

Thus, you must evaluate :

\begin{align}\mathsf P(E\mid A)&=\dfrac{\mathsf P(E)}{\mathsf P(A)}\\[2ex]&= \dfrac{\mathsf P(E,G_1)+\mathsf P(E,G_2)+\mathsf P(E,G_3)}{\mathsf P(A,G_1)+\mathsf P(A,G_2)+\mathsf P(A,G_3)}\\[2ex]&=\dfrac{\mathsf P(G_1)~\mathsf P(E\mid G_1)+\mathsf P(G_2)~\mathsf P(E\mid G_2)+\mathsf P(G_3)~\mathsf P(E\mid G_3)}{\mathsf P(G_1)~\mathsf P(A\mid G_1)+\mathsf P(G_2)~\mathsf P(A\mid G_2)+\mathsf P(G_3)~\mathsf P(A\mid G_3)}\\[2ex]&~~\vdots\\[2ex]&=\phantom{\tfrac 7{12}}\end{align}

Condition on whether the girl who opened the door is the eldest child overall, the middle child, or the youngest child. Each of these occur with probability $$\frac{1}{3}$$. We are told this within the problem statement.

The probability she is the eldest girl given that she is the eldest child is clearly $$1$$.

The probability that she is the eldest girl given that she is the middle child is the same as the probability the eldest child is a boy which is $$\frac{1}{2}$$.

The probability that she is the eldest girl given that she is the youngest child is the same as the probability that the two eldest children are both boys which is $$\frac{1}{4}$$.

This makes the overall probability:

$$\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{4}\right) = \frac{7}{12}$$