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There are $2$ boys and unknown number of girls in a nursery. A new baby is just born inside the room. We pick randomly a baby from the room, it turns out that the baby is a boy. What is the probability that the new baby just born is a boy?

We can solve this using Bayes' rule, as follows, $P(\text{new baby is boy} | \text{picked a boy})$. We end up getting $\frac{3}{5}$. Why is this answer intuitively not dependent on the number of girls in the nursery?

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  • $\begingroup$ For the answer, it is important to specify if you pick newborn babies with or without returning them after having determined their gender. $\endgroup$
    – NeitherNor
    Jul 27, 2020 at 17:45
  • $\begingroup$ @NeitherNor what do you mean by returning the baby? $\endgroup$
    – Radial Arm Saw
    Jul 27, 2020 at 18:35
  • $\begingroup$ @Radial I mean, do you pick the random baby from the nursery with or without replacement, i.e. are you returning the baby afterwards or do you keep it? $\endgroup$
    – NeitherNor
    Jul 27, 2020 at 20:24
  • $\begingroup$ @NeitherNor I don’t quite follow what you’re saying. If you’re asking if we pick multiple times, then the answer is that we only pick once (As the problem does not state we get a second pick). $\endgroup$
    – Radial Arm Saw
    Jul 27, 2020 at 20:37
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    $\begingroup$ @NeitherNor If we only pick once then it doesn’t matter whether or not the baby stays in the nursery. $\endgroup$
    – Radial Arm Saw
    Jul 27, 2020 at 20:58

2 Answers 2

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Before the new baby is born, there are 2 boys (Andy and Bob). The new baby (NB) is born, then we pick a baby at random and get a boy. In order for this to happen, one of the following scenarios occurred:

  1. NB is a boy and we picked Andy

  2. NB is a boy and we picked Bob

  3. NB is a boy and we picked NB

  4. NB is a girl and we picked Andy

  5. NB is a girl and we picked Bob

In three of these five cases, NB is a boy, so the probability is $\frac{3}{5}$

The number of girls at the start is irrelevant, because changing that number doesn't change the relative probability of any of scenarios $1-5$ occurring

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    $\begingroup$ "The number of girls at the start is irrelevant, because changing that number doesn't change the probability of any of scenarios 1−5 occuring" - probably better to say that it doesn't change that the scenarios are equally likely (which would probably also be good to point out explicitly). Adding more girls does lower the probability of scenarios 1 - 5, just not in a way that's meaningful to the result. $\endgroup$
    – Milo Brandt
    Jul 28, 2020 at 3:01
  • $\begingroup$ Yes. That's what i meant $\endgroup$
    – Moko19
    Jul 28, 2020 at 15:06
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Suppose there are initially $g$ girls in the room. Let $A$ be the event that the new baby is a boy, $B$ the event that the baby picked is a boy. I presume we are supposed to assume $\mathbb P(A) = 1/2$ (although in real life that is not quite true). Then $\mathbb P(B\mid A) = 3/(g+3)$ and $\mathbb P(B\mid A^c) = 2/(g+3)$, so $$\mathbb P(A\mid B) = \frac{\mathbb P(B\mid A) \mathbb P(A)}{\mathbb P(B\mid A) \mathbb P(A) + \mathbb P(B\mid A^c) \mathbb P(A^c)} = \frac{3}{5}$$ The point is that $g$ has the same effect on both $\mathbb P(B\mid A)$ and $\mathbb P(B\mid A^c)$, a multiplicative factor of $1/(g+3)$ in both cases, which cancels out in $\mathbb P(A\mid B)$.

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    $\begingroup$ "in real life that is not quite true" $${}$$ The other thing that in real life is not quite true that is almost always assumed in exercises in elementary probability courses is that the sexes of babies born to a couple are independent of each other. Someone published something based on data from the 1970 census of the United States about this question and I seem to recall it left a lot of questions unanswered, and I wonder if anything further has been done. $\endgroup$
    – Michael Hardy
    Jul 27, 2020 at 18:19
  • $\begingroup$ The accepted answer makes no assumption on P(A)=0.5. So I assume it's possible using Bayes theory to also come to the same result? $\endgroup$
    – David Waterworth
    Jul 28, 2020 at 4:07
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    $\begingroup$ The accepted answer does assume $\mathbb P(A)=1/2$ by taking the five cases to have equal probabilities. $\endgroup$
    – Robert Israel
    Jul 28, 2020 at 13:19

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