# Probability exercise

A couple has $3$ kids, and $1$ of them is known to be male.

What is the probability that $1$ (only $1$) of the other $2$ kids is male?

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Try writing out a list of all the possible combinations of males and females. Although the answer does depend on how it is known one of them is male. If one of the 3 kids is picked randomly, and is male, that's different. – SiliconCelery Jul 29 '12 at 10:07
I'm not sure how to consider the knowledge of "already" having a male. Seems to me that the answer is 1/2, because the events on the "other" $2$ kids is independent from the event of "already" having $1$ male. Is that right? – Adam Jul 29 '12 at 10:10
If we know which of the 3 is male, then there are 4 possible configurations for the remaining 2, in which case 2 of the 4 possible configurations meet our requirements (only one other male). If, however, we only know that one child is male (but do not know which one) there are 7 possible configurations, of which 3 of them are valid. – David Wheeler Jul 29 '12 at 10:24
I agree with your answer. But I think the solution is $1/2$ because the question says the "other $2$", meaning that we just focus on them $2$. PS: I found this question in a book, so I'm just reporting it as it is formulated. – Adam Jul 29 '12 at 10:27
@Adam you are correct,because from two child,each one is independent to each other,probability is $1/2=0.5$ – dato datuashvili Jul 29 '12 at 10:40

in generally,if we consider $3$ child,and we know that only $1$ is male,then probability of being any choosen child male is $1/3$,but if our space is consisted by two child then possibility of all combination is,let choose male as $M$,and female $F$,so $(F,F)$,$(F,M)$,$(M,F)$,$(M,M)$ probability is $2/4=1/2$

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I'll give the answer for a slightly easier variation and let you work out the rest for yourself.

Question : A family has two kids, at least one of which is a boy. What is the probability that the other kid is also a boy ?

Let A be the event "at least one of the kids is a boy" and B the event "both kids are boys". We want to know the conditional probability $P_A(B)$, so we write $$P_A(B) = \frac{P(A \cap B)}{P(A)} = \frac{P(B)}{P(A)} = \frac{\tfrac{1}{4}}{\tfrac{3}{4}} = \frac{1}{3}$$

A similar reasoning gives the answer you're looking for : $\frac{3}{7}$ (not $\frac{1}{2}$).

Edit

There seems to be quite a lot of confusion on this issue, so I'll give more details.

We consider a family of three kids, and we make the (reasonable) assumptions that each kid has a probability $\frac{1}{2}$ of being male, independently of all the others. The question (rephrased for clarification) is "If we know that there is at least one boy in a family, what is the probability that there are exactly two boys in that family?".

The important thing to understand is that this is a conditional probability. More precisely, we want to calculate $P(B|A)$, where

• $A$ : "there is at least one boy in the family"
• $B$ : "there are exactly two boys in the family"

The formula for $P(A|B)$ is $P(B|A) = \frac{P(A \cap B)}{P(A)}$. Here, we have $P(A \cap B) = P(B)$ (if there are two boys in the family, there is necessarily at least one boy). Thus, we simply need the values of $P(A)$ and $P(B)$.

If we know what random variables are, we can simply observe that the number of boys in the family follows a binomial distribution $\mathcal{B}(3,\frac{1}{2})$, but this is not really needed. We can simply list all the possible outcomes of the experiment and count how many of these outcomes correspond to our events. In order to do that, we need to ensure that all the outcomes we define have the same probability. This will be the case if we define our outcomes to be tuples (ordered lists) of the sexes of the children. We then have 8 outcomes : $$\Omega = \{(M,M,M),(M,M,F),(M,F,M),(M,F,F),(F,M,M),(F,M,F),(F,F,M),(F,F,F)\}$$ 7 of these outcomes correspond to $A$ (all but the last one), 3 correspond to $B$.

So we get $P(A)=\frac{7}{8}$, $P(B)=\frac{3}{8}$ and $\boxed{P(B|A)= \frac{\tfrac{3}{8}}{\tfrac{7}{8}}=\frac{3}{7}}$, which is the desired answer.

Alternatively, we could say that there are only seven possible outcomes $$\Omega' = \{(M,M,M),(M,M,F),(M,F,M),(M,F,F),(F,M,M),(F,M,F),(F,F,M)\}$$ each of which has probability $\frac{1}{7}$ and three of which correspond to the event $B$. We get the same answer, this time by changing our universe (instead of using the usual technique of keeping the same universe and using conditional probabilities).

EDIT : last update, I'm only supposed to teach this during the school year :) This is an answer to a comment below by @Adam which I could not fit in a comment.

The first thing to understand is that the number of possible outcomes is not important. Actually, different (but equally valid) ways to model your situation will give you different universes and different numbers of possible outcomes. That is not a problem because the probabilities of these outcomes will also be different so you will get the same probability for all your events in the end (assuming your models are both valid, obviously).

Second thing : if the question was "a family has two kids, their first-born is a boy, what is the probability that the other is a boy ?", the answer would be $\frac{1}{2}$. So the whole thing depends on how we interpret the "one of which is known to be male" in your original question (which is extremely ambiguous, almost certainly in purpose). A few things to think about :

• you meet a man, he tells you he has two siblings. What is the probability that he has exactly one brother ? Answer : $\frac{1}{2}$.
• you meet a man, he tells you he has three kids. You ask him whether they're all girls, he answers no. What is the probability that he has exactly two sons ? Answer : $\frac{3}{7}$

If you understand the reasoning behind these answers, then you have understood whatever it was that exercise was supposed to teach you.

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i have question .if we know that at least one of them is boy.we know that for second child,possibility is boy or girl,because second child does not depend on first one,it means that both have equaly likely possibility or $1/2$ – dato datuashvili Jul 29 '12 at 10:45
@dato Actually, it's quite tricky : if we know that the eldest is a boy (for instance), then the probability that the youngest is a boy is $\frac{1}{5}$. If we know that at least one is a boy, the probability that the other is a boy is only $\frac{1}{3}$. Basically, if you know that at least one is a boy, your universe is $\{(M,F),(F,M),(M,M)\}$ (and the probability is uniform). Exactly one of those three possibilities corresponds to the event "the other is a boy", so the answer is indeed $\frac{1}{3}$. – Jbeuh Jul 29 '12 at 10:50
but look for this example,we dont need write $(M,F)$ twice yes?they are same – dato datuashvili Jul 29 '12 at 10:53
No they're not ! If you consider the outcomes of two flips of a (balanced) coin and you are only interested in the number of "heads" you get, you can use $\{(H,H),(H,T),(T,T)\}$ as your universe, but in that case you do not have equiprobability : the outcome $(H,T)$ is twice as likely as $(T,T)$. – Jbeuh Jul 29 '12 at 10:57
ok thanks very much – dato datuashvili Jul 29 '12 at 10:57