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?
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.
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.
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*}
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.
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*}
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.)
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.
– karlahrnndz Jun 16 '16 at 16:10Conditional 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.
$A$ 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 $\frac{1/32}{1/16}=\frac12$.
Notice that event $A$ is independent on event $B$.
Therefore, $P(A)=\frac12$.
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.
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.
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 :
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.
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=$\frac{1}{2}$
