Probability that Your Wife Had a Boy

You may have noticed, but I have been on a brainteaser kick; I blame it on reading online on how to prepare for interviews–I’m trying to get a job! I found this probability question on Quora.

Suppose your wife is at the hospital, and she’s just given birth to a health baby; but you do not know if it’s a boy or girl. The baby has been taken to the hospital nursery, where before the addition of your infant, there are 2 boys and an unknown (but fixed) number of girls n\in\mathbb{N}. The nurse selects an infant at random from the nursery. Given that the nurse’s selection is a boy, what is the probability that your wife gave birth to a boy?

Proof. The problem can be solved with Bayes’ theorem. Let I_{1} and I_{2} be the Bernoulli random variables representing the sex of your baby and the sex of the baby the nurse picks up from the nursery, respectively. We assume that a woman is equally likely to have a boy as a girl, so \mathbb{P}(I_{1}=1)=\mathbb{P}(I_{1}=0)=\frac{1}{2}. We are interested in the probability that I_{1}=1 given I_{2}=1, which we write as

\displaystyle\mathbb{P}(I_{1}=1\mid I_{2}=1)

By Bayes’ theorem,

\displaystyle\mathbb{P}(I_{1}=1\mid I_{2}=1)=\frac{\mathbb{P}(I_{2}=1\mid I_{1}=1)\mathbb{P}(I_{1}=1)}{\mathbb{P}(I_{2}=1)}=\frac{1}{2}\frac{\mathbb{P}(I_{2}=1\mid I_{1}=1)}{\mathbb{P}(I_{2}=1)}

There are n girls and 3 boys in the nursery, if the wife has a boy. Since we assume that each infant in the nursery has equal probability of being selected by the nurse, we see that

\displaystyle\mathbb{P}(I_{2}=1\mid I_{1}=1)=\frac{3}{n+3}

To compute \mathbb{P}(I_{2}=1), we condition on I_{1}, obtaining

\begin{array}{lcl}\displaystyle\mathbb{P}(I_{2}=1)&=&\displaystyle\mathbb{P}(I_{2}=1\mid I_{1}=1)\mathbb{P}(I_{1}=1)+\mathbb{P}(I_{2}=1\mid I_{1}=0)\mathbb{P}(I_{1}=0)\\[.7 em]&=&\displaystyle\frac{1}{2}\left[\mathbb{P}(I_{2}=1\mid I_{1}=1)+\mathbb{P}(I_{2}=1\mid I_{1}=0)\right]\\[.7 em]&=&\displaystyle\frac{1}{2}\left[\frac{3}{n+3}+\frac{2}{n+3}\right]\\[.7 em]&=&\displaystyle\frac{5}{2(n+3)}\end{array}

Substituting these results in, we obtain

\displaystyle\mathbb{P}(I_{1}=1\mid I_{2}=1)=\frac{1}{2}\dfrac{\frac{3}{n+3}}{\frac{5}{2(n+3)}}=\dfrac{3}{5}

Your wife had a boy with probability \frac{3}{5}, given the information that the nurse selected a boy at random from the nursery.

This entry was posted in math.PR, Problem Solving and tagged , , . Bookmark the permalink.

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