Suppose that from an arbitrary family of $3$ children we choose one of them at random. If the chosen one is a girl, what is the probability that she has a sister and a brother?

(Note: suppose that the probability of being a boy or girl is the same and that the events are independent. i.e the sex of one child doesn't affect the probability of the sex of the other ones.)

What I have tried:

So this is probably wrong, but my approach was saying this is the same as asking what's the probability of two of them being girls and the other one a boy. What is really bugging me is that BGG, GGB, and GBG appear to be the same, so I don't really know what the sample space is...

So my next approach was saying, this can be rephrased as saying: $$P(2\text{ girls, }1\text{ boy)}=P(2\text{ girls})-P(3\text{ girls})$$ So $P(2\text{ girls})$ must be equal to $P(2\text{ boys})$ since any family has at least 2 boys or 2 girls and neither can happen in the same family, meaning $P(2\text{ girls}) + P(2\text{ boys})=1$, then $P(2\text{ girls})=\frac{1}{2}$. $P(3\text{ girls})$ must be $\frac{1}{8}$ and so $P(2\text{ girls, }1\text{ boy})= \frac{1}{2} - \frac{1}{8} = \frac{3}{8}$

Is this correct or am I misunderstanding the problem? Or could it be $P(1\text{ boy, }2\text{ girls}|1\text{ girl})$ ? Thanks in advance.

Oh, and the problem says you can use Bayes to solve this but I dont see how to be honest...


You're probably making this way too complicated for yourself. If we draw a tree diagram:

   G       B
 G   B   G   B

Since we know the first chosen is a girl, it reduces to this:

 G   B

Each of these probabilities is $1/4$ and we want either GB or BG so the probability is simply $1/2$.

Or, using Bayes' theorem:

$$ P(GB|G) = \frac{P(G|GB) P(GB)}{P(G)} $$

We know since the probability of girl/boy is independent, then $P(G|GB) = P(G)$. So this just reduces down to

$$ P(GB|G) = \frac{P(G) P(GB)}{P(G)} = P(GB) = \frac{1}{2} $$

  • $\begingroup$ Isn't P(GB)=P(G)P(B) = frac{1}{4} ? $\endgroup$ – StackeandoAndo Feb 21 '19 at 4:22
  • $\begingroup$ @ManuelPico $P(GB)$ refers to the unordered pair GB, so it is $P(GB)=P(G)P(B)+P(B)P(G)$ $\endgroup$ – Daniel Mathias Feb 21 '19 at 4:31
  • $\begingroup$ @Mathias Oh thanks! $\endgroup$ – StackeandoAndo Feb 21 '19 at 5:02
  • 1
    $\begingroup$ If you let $A$ be the event that there are two girls and a boy (in any order), $C$ be the event that the first child you meet is a girl, then you can change each $GB$ to $A$ and each singleton $G$ to $C$ in the Bayes's Theorem portion of this answer and not have the ambiguity about whether $GB$ implies a birth order. $\endgroup$ – David K Feb 21 '19 at 5:02
  • $\begingroup$ We know the chosen child is a girl; we do not know that she is first. $\endgroup$ – Graham Kemp Feb 21 '19 at 5:05

There are $2^3$ outcomes in the preliminary sample space $\{B,G\}^3$, and these may be assumed to be reasonably equally probable. That is eight. We then select the first, second, or third child without bias, and it so happens that we do select a girl.

Intuitively, take eight boxes and place three blue-or-green marbles in each as: one box with three green marbles, three boxes with two green marbles, three boxes with one green marble, and the last box has no green marbles.

There are $3+6+3+0$ green marbles, from which we may select without bias, of which $6$ are in boxes with two green and one blue marble.

Mathematically, let $G$ be the count of girls, and $S$ the event of selecting a girl.   Then by conditional probability and the law of total probability:

$$\begin{align}\mathsf P(G=2\mid S)&=\dfrac{\mathsf P(S\cap G=2)}{\mathsf P(S)}\\[1ex]&=\dfrac{\mathsf P(S\cap G=2)}{\mathsf P(S\cap G=1)+\mathsf P(S\cap G=2)+\mathsf P(S\cap G=3)}\\[1ex]&=\dfrac{\tfrac 23\binom 32\tfrac 18}{\tfrac 13\binom 31\tfrac 18+\tfrac 23\binom 32\tfrac 18+\tfrac 33\binom 33\tfrac 18}\\[1ex]&=\dfrac{6}{3+6+3}\\[1ex]&=\dfrac{1}{2}\end{align}$$

  • $\begingroup$ Aren't we talking about the probability she has a brother and a sister? You seem to have included the possibility of [girl, girl, girl] as well. $\endgroup$ – Infiaria Feb 21 '19 at 5:40
  • $\begingroup$ Right. Updated. $\endgroup$ – Graham Kemp Feb 21 '19 at 9:44

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