Finding probability of other child also being a boy A man visits a couple who have two children.One of them, a boy, comes in to the room. Find the probability p that other is also a boy
a) 1/3 b) 2/3 c) 1/2 d) 3/4
Correct Ans: a) 1/3
However I am getting answer as 1/2.My solution is as follows.
The couple has two children, hence the sample space is as follows.
B- Boy G - Girl
BB BG GG ( BG and GB mean the same thing as Girl,Boy), but it is already known that one is boy.So the sample space reduces to
BB BG
So now p is 1/2
Is there some mistake is my approach ?
 A: This kind of problem is tricky. I would strongly advise making an explicit probability model, so that all assumptions are clear. We assume that children come in order into a "family," and that their sexes are given by successive flips of a fair coin. (That model is certainly not quite correct.) We will also assume that the two children flip a fair coin to determine who will be first to greet the visitors. In real families, the loser probably has to go first. 
Let $X$ be the event that a boy comes first into the room, and let $T$ be the event the family is a two-boy one. We want $\Pr(T|X)$. By the usual conditional probability formula, we have
$$\Pr(T|X)\Pr(X)=\Pr(T\cap X).$$
It remains to calculate $\Pr(X)$ and $\Pr(T\cap X)$.
The event $X$ can happen in two ways: (i) It is a two-boy family or (ii) It is a one-boy family, and the boy lost and had to go  first.
Easily, the probability of (i) is $1/4$. For (ii), the probability we have a mixed family is $2/4$. Given that it is a mixed family, the probability the boy went into the room first is $1/2$. So the probability of (ii) is $(2/4)(1/2)$. 
We conclude that $\Pr(X)=\dfrac{1}{4}+\dfrac{2}{4}\cdot\dfrac{1}{2}=\dfrac{1}{2}$. 
We have already calculated $\Pr(T\cap X)$: it is just the probability of (i), which is $\dfrac{1}{4}$. 
It follows that $\Pr(T|X)=\dfrac{\Pr(T\cap X)}{\Pr(X)}=\dfrac{\frac{1}{4}}{\frac{1}{2}}=\dfrac{1}{2}$.
A: Assume that boys and girls are born in equal overall proportions, that every boy is curious about visitors and soon comes to see them when they enter the house, but that every girl is busy studying mathematics, pour l'honneur de l'esprit humain. As a consequence, girls do not care about such mundane matters as visitors coming to their house and they never stop their work to come to see them. 
Then the probability that the other child is a boy is one-third.
Assume now that everything is as before except that the visitors are all mathematicians coming from the Institute nearby. As a consequence, girls are now as eager to come and see them than boys.
Then the probability that the other child is a boy is one-half.
Assume finally that everything is as in the second version except that boys find mathematicians more and more boring, hence they stop gradually to react to these arrivals. Meanwhile girls are as eager as ever to see and talk to these visitors.
Then the probability that the other child is a boy converges to one.
Morality: It happens that word problems are ambiguously stated.
A: This is not the same question as the one mentioned in the comment by Martin Sleziak, and your "correct ans" is wrong; the probability is indeed $\frac12$ (under the assumption, not quite true in reality but reasonable for this question, that the probability of a random person being male $\frac12$, and that it is independent of the gender of any other fixed person).
The question is equivalent to the following one: you pick a random person $p$ and ask what are his siblings; it turns out $p$ has just one brother. What is the probability that $p$ is male?  You could also say, "pick a random man $m$ with one sibling, what is the chance that his sibling is a brother" (which more closely resembles the setup of your question); it is just another equally random way to make the selection (take $m$ to be the brother of $p$).
There are all kinds of ways to see the answer is $\frac12$ here, basically because the gender of one sibling is independent of the other. If you like detail: there are $4$ possibilities for genders in $2$-child families, in oldest-youngest order $FF,FM,MF,MM$, all equally likely. Person $p$ could be oldest or youngest, treat the cases separately (they give equal probabilities in the end anyway). Supposing $p$ is oldest, the fact that the younger sibling is a brother eliminates two possibilities leaving $FM$ and $MM$; this leaves equal probabilities for $p$ being female or male. If $p$ is youngest, then $MF$ and $MM$ are left, again leaving equal probabilities for $p$ being female or male.
The essential point that distinguishes this question from the one linked to is that you are not given that "one of the siblings is a boy", which gives a different kind of information (eliminating only one of four possibilities). Here a specific sibling is found to be a boy.
A: This is a case of conditional probability: if $\,B_i\,,\,G_i\,$ denote the event that the $\,i-th\,$ kid is a boy ( a girl ) , we thus are looking for
$$P(B_2/B_1)=\frac{P(B_1\cap B_2)}{P(B_1)}=\frac{1/4}{1/2}=\frac{1}{2}$$
The above, of course, under the mild though usually non-true assumption that some random couple's kids have the very same probability to be boys as to be girls.
A: Assume that every couple keeps having children until they have a son and then they stop.  Furthermore assume that the probability of multiple simultaneous births (aka twins, triplets, etc.) is zero.
Then the probability is 0% that the other child is a boy.
With these assumptions, we can also tell that the two children are older sister and younger brother.
A: There are four (4) equally likely events in the sample space:
{BB, BG, GG, GB}  (Yes, BG and GB are different).
If one of the children is a boy, then GG cannot possibly be the case. This leaves {BB, BG, GB}.
P(BB) = 1/3.
A: Possibilities are: MF FM MM FF
Eliminate FF, we know one is a boy
left with MM, MF, FM
FM is the same as MF
1/3 chance MM
2/3 chance MF
