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A family has two children. Find the probability that both are girls given that at least one one is a winter-born girl.

I understand all the steps in the solution below except one: "To compute the numerator, use the fact that "both girls, at least one winter girl" is the same event as "both girls, at least one winter child." How does this hold? Doesn't conditioning on knowing that one is a girl make the probability that both are girls higher than the case in which we just condition on season?

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beginner
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  • For what it's worth, much of my confusion centered on the related question: what is the probability of two girls, given that there is at least one girl. I regard the answer to this as $(1/3)$. Further, by symmetry, if you accept the answer of $(7/15)$, then that answer must also apply if you are given that there is at least one summer girl, or at least one fall girl, or at least one spring girl. But, the combination of these 4 seasonal questions seem to equate to the probability of at least one girl, which seems to have probability of $(1/3)$. Thus, my confusion, so I deleted my answer – user2661923 Aug 01 '21 at 20:10
  • Re previous comment: suppose you accept all of it's math. Then, the problem of : probability of two girls, given at least one girl, may be attacked as follows: either the one known girl was born in the summer, fall, spring, or winter. In each of those 4 possibilities, the chance that both children are girls has been computed as $(7/15)$. Therefore, the probability of two girls, given at least one girl is $(7/15)$. This seems to contradict the answer of $(1/3)$. Therefore, either the analysis in this comment is (somehow) invalid, or the analysis in the question is invalid. – user2661923 Aug 01 '21 at 20:25
  • Re previous comment, my guess, which could easily be wrong, is that the analysis in the previous comment is invalid. That is, I am guessing that the probability of two girls given at least one is a girl, is somehow different than the probability of two girls given that one is a winter girl. That is, the probability of two girls given that one is a girl is different from the probability of two girls given that the oldest is a girl. I am on very shaky ground here, and I hope that a Probability expert will weigh in. – user2661923 Aug 01 '21 at 20:31
  • @user2661923 I agree with the first probability of $1/3$, that is if we are given one of them is a girl, then the probability is $1/3$ that both are girls. But your second part does not seem correct. If we go by that logic, conditional probability of both being girls given at least one of them is a winter girl should be lower than $1/3$ but even if you do not agree with $7/15$, why would that be lower than $1/3$? – Math Lover Aug 01 '21 at 20:39
  • @MathLover I am not disagreeing with you, but am instead simply expressing confusion. I (also) suspect that my analysis is invalid, but am not sure why. Please explain why my logic would imply that if you are given that one of them is a winter girl, then the chances that both are girls is less than (1/3). Or, if you can, please explain the flaw (which I suspect is there) in my analysis. If your response is (necessarily) long-winded, perhaps you might edit your answer to include your response to this comment. – user2661923 Aug 01 '21 at 20:45
  • @user2661923 I said less than $1/3$ based on your second comment where you mentioned symmetry. I think because we have a given condition that at least one of them is a winter girl which is less probable event than at at least one girl, makes mathematically more likely that the second child is a girl. – Math Lover Aug 01 '21 at 20:56
  • Does this answer your question? In a family with two children, what are the chances, if one of the children is a girl, that both children are girls? – Peter O. Aug 01 '21 at 22:30
  • @PeterO. Personally, on an intuitive level, the question of why the answer to the posted question equals $(7/15) \neq (1/3)$ still persists. – user2661923 Aug 01 '21 at 22:59
  • @user2661923 I think you go wrong here: "either the one known girl was born in the summer, fall, spring, or winter." There is no "the one known girl" given at least two girls: there is either one girl or (less likely) two, and in the case of two girls it is possible that one was born in summer and one in the fall, and likewise with any other pair of seasons. That is, it is possible that the conditions "at least one winter girl" and "at least one summer girl" are both true simultaneously, whereas you are trying to combine the winter and summer cases as if they were mutually exclusive. – David K Aug 02 '21 at 01:04
  • @user2661923 If that does not resolve the paradox for you, you might try asking your question on its own. Your point of confusion seems to be quite different from that of the question above. – David K Aug 02 '21 at 01:13

3 Answers3

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"Both girls, one winter girl" means that both children are girls, and out of those two girls, one of them must be a winter girl.

"Both girls, one winter child" means that both children are girls, and out of those two girls, one of them must be a winter child, irrespective of whether that winter child is a boy or a girl.

In the numerator, both children have to be girls, and the winter child is one of those two children, so the winter child must be a girl.

Therefore, the two situations are equivalent.

I hope this explanation helped; if you have any more doubts, feel free to comment on this answer.

Saksham Sethi
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This is one of those problems that is susceptible to a simple table of events.

In the table below, the row labels on the left are the possible events for the first child: $G_1$ if a girl, $B_1$ if a boy, $W_1$ born in winter, $P_1$ born in spring, $S_1$ born in summer, $F_1$ born in the fall. The column labels are the possible events for the second child, using the same letters as labels but with the subscript $2.$

Each of the 64 cells of the table under the column labels has (we assume) equal probability, since we assume male and female births equally likely, births in each season equally likely, season and sex are independent, and the births of the two children are independent.

All cells that satisfy "both girls, at least one winter girl" are labeled $A.$ The exact same set of cells satisfy "both girls, at least one winter child": there is no other cell where both children are girls and at least one child was born in the winter. If you doubt this, check all the other cells and see whether any of them is an event in which both children are girls and at least one child was born in the winter.

\begin{array}{c|cccccccc} {} & G_2W_2 & G_2P_2 & G_2S_2 & G_2F_2 & B_2W_2 & B_2P_2 & B_2S_2 & B_2F_2 \\ \hline G_1W_1 & A & A & A & A & C & C & C & C \\ G_1P_1 & A & D & D & D & E & E & E & E \\ G_1S_1 & A & D & D & D & E & E & E & E \\ G_1F_1 & A & D & D & D & E & E & E & E \\ B_1W_1 & C & E & E & E & F & F & F & F \\ B_1P_1 & C & E & E & E & F & F & F & F \\ B_1S_1 & C & E & E & E & F & F & F & F \\ B_1F_1 & C & E & E & E & F & F & F & F \end{array}

If the event $A$ comprises exactly the cells labeled $A,$ event $C$ comprises exactly the cells labeled $C,$ and so forth, we have

\begin{align} A & = \text{both girls, at least one winter girl;} \\ A \cup C & = \text{at least one winter girl;} \\ \end{align}

and by simple counting it is easy to confirm that $P(A) = 7/64$ and $P(A\cup C) = 15/64.$

All the other calculations in the given solution are easily checked as well, although it is not really necessary to make a table to do so. For example, one of the implicit steps in the given solution is

$$ \mathsf P(\text{both girls, at least one winter child}) = \mathsf P(\text{both girls}) \mathsf P(\text{at least one winter child}), $$

which follows directly from the assumption that the seasons of birth are independent of the sexes of the children. (In the next step we substitute $\mathsf P(\text{both girls}) = 1/4,$ and that step is explicitly shown in the solution.)

For comparison, the classic two-girls problem hinges on these events:

\begin{align} A \cup D & = \text{both girls;} \\ A \cup D \cup C \cup E & = \text{at least one girl.} \\ \end{align}

Taking away the "winter" condition allows some additional probability of two girls (cells labeled $D$), but it also allows a lot more additional probability of just one girl (cells labeled $E$).

David K
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You are applying Bayes' theorem.

$P(A|B) = \cfrac{P(A \cap B)}{P(B)}$

$A$ is the event of both children being girls and $B$ is the event of at least one of the children being winter girl.

Now if $C$ is the event of at least one of them being winter child, $P(A \cap B) = P(A \cap C)$ as $B \subseteq C$ and $|A \cap (C-B)| = \emptyset$.

You can always find $P(A \cap B)$ directly too.

Probability if both are winter girls: $\cfrac{1}{8^2}$

Probability if exactly one of them is a winter girl and the other is a non-winter girl: $2 \cdot \cfrac{1}{8} \cdot \cfrac{3}{8}$

Adding them up, we have $\cfrac{7}{64}$ which is same as the official solution.

Math Lover
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