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A and B are married. They have two kids. One of them is a girl. What is the probability that the other kid is also a girl?

Someone says $\frac{1}{2}$, someone says $\frac{1}{3}$. Which is correct?

Now A and B have 4 children and all of them are boys. B is pregnant. So what is the probability that A and B are gifted with a baby girl? Is it $\frac{1}{2}$ or there will be some conditional probability?

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    $\begingroup$ The first question comes up a lot...here for example. For the second, you need to be more clear. Are you asking a practical question about the independence of gender statistics in families or an abstract question assuming that one child's gender is independent of all the others? If the latter, then $\frac 12$. $\endgroup$ – lulu May 26 '16 at 10:46
  • $\begingroup$ Are you assuming that each individual birth has a 50/50 chance of being male or female? Or are you allowing for the fact that various sex selection factors act at conception, or between conception and birth, so that having 4 boys could indicate one of the partners has some factor tending to produce boys? $\endgroup$ – Colin McLarty May 26 '16 at 10:49
  • $\begingroup$ No biological preference is assumed,50/50 chance for male/female $\endgroup$ – Hailey May 26 '16 at 10:50
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Regarding the first question:

A and B are married. They have two kids. One of them is a girl. What is the probability that the other kid is also a girl?

This is a classical problem in probability. What makes this question confusing is the fact that the question is ambiguous since there may be no such thing as "the other kid".

Suppose a couple has two children, call then $\mathcal{X}$ and $\mathcal{Y}$. When you say "one of them is a girl" two possible interpretations are:

#1 Someone told me the couple has at least one girl but I don't know if both children are girls, if $\mathcal{X}$ is a girl, or if $\mathcal{Y}$ is a girl; the possible cases are GG, GB, BG. I want to compute the probability that both of the couple's children are girls (GG).

or

#2 Someone told me $\mathcal{X}$ is a girl but I don't know if $\mathcal{Y}$ is a boy or a girl; the possible cases are GG, GB. I want to compute the probability that $\mathcal{Y}$ is also a girl (GG).

Now, you can answer each of these questions using, for example, expressions for conditional probabilities (the answers to interpretations #1 and #2 are 1/3 and 1/2, respectively). However, if you are left with a feeling of "sure, I see how you can obtain that result but why isn't my answer correct?" read on.


Interpretation #1:

(This is the one I find interestingly confusing.)

When you first think about the problem you might use the following (incorrect) reasoning:

I know at least one of the children is a girl, what is the probability that the other child (the child of unknown gender) is also a girl? well, shouldn't it be 1/2? I mean, the other child can be a boy or a girl regardless... right?

One of the problems with this reasoning is that there is no such thing as "the child of unknown gender" (therefore, there is no such thing as "the other child"). In fact, you do not know either of the children's genders (in this interpretation both children have unknown gender!). All you know is that there is a girl in there somewhere, but you do not know which of the two children is the girl. How can you hope to compute the probability that the other child (the child of unknown gender) is a girl if there is no such thing as "the child of unknown gender"? In fact, if you happen to know that $\mathcal{X}$ (or $\mathcal{Y}$) is a girl then you are using a different interpretation of the question (interpretation #2).

The other problem with this reasoning is related to independence. As people have pointed out, given the added overarching assumption that there is at least one girl, "$\mathcal{X}$'s gender" and "$\mathcal{Y}$'s gender" are not independent random variables. This is easy to see since, for example, if $\mathcal{X}$ is a boy then $\mathcal{Y}$ is forced to be a girl with probability 1 (not 1/2). This lack of independence is the reason why you can't say that one of the two children is a boy or girl independently/regardless of the other.

Formal Derivation of Answer to Interpretation #1:

Using the formula for conditional probabilities we have \begin{align*} P\big(\text{Both are Girls } \big| \text{ At Least One Girl}\big)&=\frac{P\Big(\big(\text{Both are Girls}\big)\cap\big(\text{At Least One Girl}\big)\Big)}{P\big(\text{At Least One Girl}\big)}\\ &=\frac{P\big(\text{Both are Girls}\big)}{P\big(\text{At Least One Girl}\big)}\\ &= \frac{1/4}{3/4}=\frac{1}{3}. \end{align*}


Interpretation #2:

I don't think there is anything confusing about this interpretation. You know the $\mathcal{X}$ is a girl and $\mathcal{Y}$ can be a girl or a boy independently of $\mathcal{X}$. In this case, the random variables "$\mathcal{X}$'s gender" and "$\mathcal{Y}$'s gender" are not linked by any added overarching assumption and are independent.

Formal Derivation of Answer to Interpretation #2:

Using the independence of the second child's gender relative to the first child's gender:

\begin{align*} P\big(\text{Second is a Girl } \big| {\text{ First is a Girl}}\big)=P\big(\text{Second is a Girl}\big)=\frac{1}{2}. \end{align*}


Regarding the second question:

Now A and B have 4 children and all of them are boys. B is pregnant. So what is the probability that A and B are gifted with a baby girl? Is it 1/2 or there will be some conditional probability?

The answer is that the next child's gender is independent of the first four children's genders (by assumption), this is because there is no added overarching assumption involving the next child. The answer is then simply $1/2$. (This is just like in Interpretation #2.)

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  • $\begingroup$ +1 Thanks for the lovely answer to an often asked question. I know that question has been asked in some form on this site in more than one place already. Whenever someone finds one of those places they should link to this. $\endgroup$ – Ethan Bolker Jun 5 '16 at 18:36
  • $\begingroup$ Thanks @EthanBolker! While I've known the mathematical answer to this problem and have been familiar with the ambiguity of the question for a while, I had always struggled to explain (without math) why the incorrect reasoning was incorrect. In my head the answer was often just "that's not how probability works". I'm glad to have come up with a semi-satisfying non-math-based explanation. $\endgroup$ – karlahrnndz Jun 5 '16 at 18:50
  • $\begingroup$ i dont see how is this reliable? i know the answer is right but P(A/B) is calculated only if A and B are dependent, how the events of giving birth of two babies are affecting eachother conditionally ?if someone roll a dice and makes 6, the second toss would never depend on the first. $\endgroup$ – Abr001am Jun 5 '16 at 19:54
  • $\begingroup$ First: you can compute P(A|B) regardless of whether A and B are dependent or not. Second: You are almost right, without other assumptions two babies can be assumed to have independent gender. However, in this case you are adding another assumption: you can't have two boys. Originally the babies genders are independent but once you start adding assumptions like "you can't have two boys" or "the second baby has to have different gender than the first baby" then the genders can become dependent random variables. $\endgroup$ – karlahrnndz Jun 16 '16 at 16:10
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Originally there are $4$ equiprobable possibilities: BB,GB,GG,BG.

Condition "at least one of them is a girl" leaves open $3$ (again) equiprobable possibilities: GB,GG,BG.

Draw your conclusion.


Answer in short on second question: independence.

Again draw your conclusion.

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Conditional probability:

Let $A$ and $B$ be two events.

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

which means, the probability of $A$ occuring given that $B$ occured, is the probability of both $A$ and $B$ occuring, divided by the probability that $B$ occurs.

In this case, $A$ is the event that the other kid is a girl, and $B$ is the event that one of the kids is a girl.

$A\cap B$ would be both kids are girls, which has a probability of $\frac14$.

$B$ would be the event that one of the kids is a girl, which has a probability of $\frac34$.

Therefore, the required probability is $\frac{1/4}{3/4}=\frac13$.


And for the second question, we can do it using conditional probability or without.

Using conditional probability:

$A$ be the event that the fifth child is a girl.

$B$ be the event that the first four children are boys.

$P(A\cap B)$ would be $\left(\frac12\right)^5=\frac1{32}$.

$P(B)$ would be $\left(\frac12\right)^4=\frac1{16}$.

The required probability would be $\frac{1/32}{1/16}=\frac12$.

Without using conditional probability:

Notice that event $A$ is independent on event $B$.

Therefore, $P(A)=\frac12$.


Notes

In the first question, the probability in question is the probability that "the other kid is also a girl", making the two events dependent.

However, in the second question, the word other is gone, leaving us with independent events.

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  • $\begingroup$ and for the second question? $\endgroup$ – Hailey May 26 '16 at 10:46
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One of the kids is a girl does not convey enough information to answer the question. This is an example where we need to be extra careful using ordinary language to express what we mean mathematically.

Does it mean :

  1. One specific kid is a girl ( for example first born ).
  2. Any one of the two kids is a girl.
  3. Exactly one of the kids is a girl (thanks to drhab's comment)

For 1 we only have the freedom left to consider two cases girl & boy which should be an equi-probable independent event. That is: probability $1/2$.

For 2 it is like sampling one of the letters $\{BB, BG, GB, GG\}$, noticing that it is a $G$ and then wondering what the other would be. As you can see there is only one G which has another G besides it but three which have any 1 G, so here the probability it would be $1/3$.

For 3 it could be if a parent says "I have two children of which one is a girl" it probably is assumed that only one of the kids is a girl and then the probability of second kid to be a girl would be 0.

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    $\begingroup$ There is even a third possibility: one of the kids is a girl could also mean exactly one of the kids is a girl. If someone informs you saying: "I have two children and one of them is a girl" then that's almost surely what is meant. Probability that the other is also a girl is $0$ then. $\endgroup$ – drhab May 27 '16 at 7:23
  • $\begingroup$ Yes great observation. I was so focussed on the question I did not even think of that. $\endgroup$ – mathreadler May 27 '16 at 11:43
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There are $4$ possibilities such that,BB,BG,GB,GG

If it is said that,one of them is girl.So,BB option is omitted.Now there are 3 possibilities.If it said another one is also girl,then it suits with GG possibility only.

So,Probability=$\frac{1}{3}$

And for second condition,if we ignore the medical statistics,then there will be 50/50 chance of having boy or girl.

So,Probability=$\frac{1}{2}$

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