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A woman planning her family considers the following schemes on the assumption that boys and girls are equally likely at each delivery:

(a) Have three children.

(b) Bear children until the first girl is born or until three are born, whichever is sooner, and then stop.

(c) Bear children until there is one of each sex or until there are three, whichever is sooner, and then stop.

Let $B_{i}$ denote the event that $i$ boys are born, and let $C$ denote the event that more girls are born than boys.

Q1. Find $\textbf{P}(B_{1})$ and $\textbf{P}(C)$ in each of the cases (a) and (b).

Q2. Find $\textbf{P}(B_{1})$ and $\textbf{P}(C)$ in case (c).

Q3. Find $\textbf{P}(B_{2})$ and $\textbf{P}(B_{3})$ in all three cases.

Q4. Let $E$ be the event that the completed family contains equal numbers of boys and girls. Find $\textbf{P}(E)$ in all three cases.

MY ATTEMPT

A1. The sample space corresponding to the (a) case is given by \begin{align*} \Omega_{a} = \{&(b,b,b),(b,b,g),(b,g,b),(g,b,b),(b,g,g),(g,b,g),(g,g,b),(g,g,g)\} \end{align*}

Therefore $\textbf{P}(B_{1}) = 3/8$ and $\textbf{P}(C) = 4/8 = 1/2$.

On the other hand, the sample space corresponding to the (b) case is given by

\begin{align*} \Omega_{b} = \{g,(b,g),(b,b,g),(b,b,b)\} \end{align*}

Considering that the probability for a girl to be born is equal to that of a boy, we get

\begin{align*} \textbf{P}(B_{1}) = \textbf{P}(\{(b,g)\}) = \frac{1}{2}\times\frac{1}{2} = \frac{1}{4}\quad\text{and}\quad\textbf{P}(\{g\}) = \frac{1}{2} \end{align*}

A2. In this case, the sample space is given by \begin{align*} \Omega_{c} = \{(b,g),(g,b),(b,b,g),(b,b,b),(g,g,b),(g,g,g)\} \end{align*}

Therefore we can express the sought probabilities as

\begin{align*} &\textbf{P}(B_{1}) = \textbf{P}(\{(b,g),(g,b),(g,g,b)\}) = \frac{1}{2}\times\frac{1}{2} + \frac{1}{2}\times\frac{1}{2} + \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2} = \frac{5}{8}\\\\ &\textbf{P}(C) = \textbf{P}(\{(g,g,b),(g,g,g)\}) = \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2} + \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2} = \frac{1}{4} \end{align*}

A3. Briefly speaking, we have $\textbf{P}_{a}(B_{2}) = 3/8$, $\textbf{P}_{b}(B_{2}) = 1/8$ and $\textbf{P}_{c}(B_{2}) = 1/8$.

Analogously, we have $\textbf{P}_{a}(B_{3}) = \textbf{P}_{b}(B_{3}) = \textbf{P}_{c}(B_{3}) = 1/8$.

A4. We have $\textbf{P}_{a}(E) = 0$, $\textbf{P}_{b}(E) = 1/4$ and $\textbf{P}_{c}(E) = 1/2$.

Firstly, is my answer correct? Secondly, is there another argument which leads to the right answer without enumerating each element from the target sample space and applying the concept of independence of events? Perhaps some symmetry argument, I don't know.

Anyway, thanks in advance for any contribution.

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For cases (a) and (c), symmetry says that you will get the same answers if you swap the words "boy" and "girl". In particular, if $G_i$ is the event of having $i$ girls, then $P(B_i) = P(G_i)$ for each $i = 0, 1, 2, 3.$

For (a) we observe that $B_i$ and $G_{(3-i)}$ are the same event, so $P(B_0) = P(G_3) = P(B_3).$ Similarly we find that $P(B_1) = P(B_2).$ You can compute that $P(B_3) = P(bbb) = \frac18,$ which means $P(B_0\cap B_3) = 2P(B_3) = \frac14,$ therefore $2P(B_1) = P(B_1 \cap B_2) = \frac34,$ therefore $P(B_1) = P(B_2) = \frac38.$

In case (a), there must be an odd number of children, therefore not equal numbers of boys and girls, so $P(E) = 0,$ and symmetry says more boys and more girls are equally likely, so $P(C) = \frac12.$ So now that's all relevant questions answered for case (a), though not in the same order they were asked.

For case (c) you get either $2$ or $3$ children. Two children implies $E$ (because that's the condition for stopping at $2$) and three children implies $E^C$ ($E$ complement or "not $E$). Whichever sex is born first, you have a $\frac12$ chance to get the other sex next, so the probability of $2$ children is $\frac12,$ that is, $P(E) = \frac12.$

In case (c), if the event $E^C$ occurs then by symmetry it's equally likely to be more boys or more girls, so $P(C) = \frac12(P(E^C)) = \frac12(1 - P(E)) = \frac14.$

In case (c) the event $B_1$ can occur in only two ways: if $E$ occurs, or if you have birth order $ggb.$ These events are disjoint. Therefore $P(B_1) = P(E) + P(ggb) = \frac12 + \frac18 = \frac58.$ Now, $B_3$ and $G_3$ are each disjoint from $B_1$ and from each other, and by symmetry $P(B_3) = P(G_3) = P(ggg) = \frac18.$ The event $B_0$ cannot occur with two children and therefore is identical to $G_3$, so the only $B_i$ event unaccounted for is $B_2$. We have $P(B_2) = 1 - P(B_0) - P(B_1) - P(B_3) = 1 - \frac18 - \frac58 - \frac18 = \frac18.$ And that's all the questions answered for case (c).

In the above there are a handful of appearances of individual elements of the sample space, but all the rest is symmetry.

Case (b) can be related the sequence of events $g, bg, bbg, bbbg, bbbbg, \ldots$ by combining all the events $bbbg, bbbbg, \ldots$ into the single event $bbb.$ This gives us event probabilities $\left(\frac12\right)^n$ for $n = 1, 2, 3$ for the first three events and $1 - \sum_{n=1}^3 \left(\frac12\right)^n = \left(\frac12\right)^3$ for the last. We could easily extend this to more than $3$ maximum children, but all the questions ask about single elements of the sample space, so it's hard to say we can do something other than enumerate each element.

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  • $\begingroup$ Well done! Thank you very much. $\endgroup$ – user1337 Jan 13 at 11:06
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Your approach is well thought out and well executed. I haven't checked all your results, but I checked enough to see that you know what you're doing.

In some of your cases a faster approach is possible; for instance, $\textbf{P}(C)$ in case (a) is obviously $1/2$ because of the boy/girl symmetry. But I don't see a better approach in general.

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