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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


So now p is 1/2

Is there some mistake is my approach ?

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marked as duplicate by M Turgeon, user127.0.0.1, TMM, T. Bongers, egreg Feb 2 at 0:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Actually your answer is right, but your approach is wrong: without any information BG is twice as likely as BB (and also as GG). –  Marc van Leeuwen Sep 13 '12 at 13:27

7 Answers 7

This is not the same question as the one mentioned in the comment by Martin Sleziak, and your "correct ans" is wrong; the probability (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) is indeed $\frac12$.

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.

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It would be different, however, if the story takes place in a culture where the norm is that male children (or children of the same gender as the guest) must greet visitors before female children do. Or even in a culture without such strict rules, if boys and girls are brought up to be interested in different things, so a boy would be more likely to find the male guest's conversation interesting, and thus more likely to come into the room. –  Henning Makholm Sep 13 '12 at 12:52
@HenningMakholm: True. And it would also be different if (avoiding an a priori boy/girl bias) one assumes that children of the same gender are likely to play together, making it unlikely that one of them would enter the room in that case. If questions like this are to make sense at all, the rule "what is not known should not be assumed" must be assumed (though that is self-contradictory...). –  Marc van Leeuwen Sep 13 '12 at 13:19
And what about "I meet a married couple, the first of which, a woman, greets me. What is the probability the other person is a man?". Also quite culture-dependent. –  Marc van Leeuwen Sep 13 '12 at 13:21
Thanks this is the perfect answer ! –  kabirkukreti Sep 14 '12 at 13:02

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.

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@vanna You should have sticked to your guns: when somebody declares This answer is incorrect, think for yourself to determine whether they are right or not. –  Did Sep 13 '12 at 18:57
hahahaha ! This is a simple probability problem not some culture specific problem ! ( pun intended please don't mind ) –  kabirkukreti Sep 14 '12 at 13:02
kabirkukreti: Not sure that I understand, nor that I agree with what I understand of, your comment. –  Did Sep 15 '12 at 4:25

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 $\Pr(T|X)=\dfrac{\Pr(T\cap X)}{\Pr(X)}=\dfrac{\frac{1}{4}}{\frac{1}{2}}=\dfrac{1}{2}$.

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Don't we usually assume apriori probability of equally likely chances of a child being either girl or boy ? –  kabirkukreti Sep 14 '12 at 13:00
@kabirkukreti: Yes, we usually do. Making assumptions explicit is a useful habit. In real situations, producing a suitable model is often the hardest part. –  André Nicolas Sep 14 '12 at 13:22

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.

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Very interesting approach.Yes they are independent events –  kabirkukreti Sep 14 '12 at 13:04

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.

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There are 4 events in the sample space. Why are they equally likely? –  emory Sep 13 '12 at 22:55
How come BG,GB are different ? –  kabirkukreti Sep 14 '12 at 13:06
@emory: We can assume that each of the two children has a 1/2 chance of being either gender (1/2 chance of being boy, 1/2 chance of being girl). There is a 1/2 chance that the first child is a boy. If the first child is a boy, then there is a 1/2 chance that the second child is either gender. By multiplication rule, this leaves a 1/4 chance for BB and a 1/4 chance for BG (1/2 $\times$ 1/2 = 1/4). Similar logic can be used to yield a 1/4 chance for GB and GG each. –  Scott Caldwell Sep 14 '12 at 16:05
@kabirkukreti: Because the two children are different. This is not necessarily intuitive, so I'll attempt to illustrate my point a similar example. Suppose you flip a fair coin twice. There are four possible (equally likely) outcomes: HH, HT, TT, TH. TH and HT are different because of the order in which the heads and tails appeared. The first coin flip has a 1/2 chance of landing either way, and the second flip, because it is completely independent of the first, also has a 1/2 chance of going either way. While not immediately obvious, this is also the case for the B/G problem. –  Scott Caldwell Sep 14 '12 at 16:21
@ScottCaldwell If we want to, we can assume that each of the two children has a 1/2 chance of being either gender and that they are independent. It helps to make these assumptions explicit. If the first child is a boy then wouldn't GB also be impossible, leaving {BB,BG} ... P(BB)=1/2. –  emory Sep 14 '12 at 23:09

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.

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Two children. $ $ –  Did Sep 13 '12 at 18:57
@did I don't understand your comment. –  emory Sep 13 '12 at 19:32
The OP assumes that each family has exactly two children, you assume they have a random positive number of children. –  Did Sep 13 '12 at 20:04
@did, The OP did not assume that each family has exactly two children. The OP specified that the visited family has exactly two children and did not say anything about other families. I assumed that the family lives in a culture where every family strives to have one son. (Sadly, some families will end up with zero children.) –  emory Sep 13 '12 at 22:50
You are absolutely right about the lack of assumptions in the question. Sorry about the noise. –  Did Sep 14 '12 at 5:02

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

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