# Question from probability theory

I have this question: A person has three children with at least one boy. Find the probability of having at least two boys among the children.

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My intuition about the problem is this-- the person has at least one boy means total possible outcome can be :

BBB,BBG,BGB,BGG,GBB,GBG,GGB

it cannot have GGG, so for at least two boy we have BBB,BBG,BGB,GBB as favorable outcome this gives us probability 4/7.

but the solution given in book goes as --->

Event = having at least two boy

The event is occurring under the following situations:

• second is a boy and third is a girl OR
• second is a girl and third is a boy OR
• second is a boy and third is a boy

so the probability will be (1/2) *(1/2) + (1/2) *(1/2) + (1/2) *(1/2) = 3/4

So please tell me is my intuition is correct or the solution given in the book

• Are all scenarios equally likely? Sep 20 '14 at 4:40
• @copper.hat yes Sep 20 '14 at 4:52
• If the question is stated the way you have asked it, then the answer is 4/7 Sep 20 '14 at 5:35
• also, while your reasoning is correct, this method will not be optimal for solving larger problems, for example if there were 5 kids instead of 3, it is worthwhile to get familiar with the binomial (and other) distributions Sep 20 '14 at 5:37

One way is to list all 8 possibilities, BBB, BBG, BGB, etc. and them remove those that don't have at least one B. Count them. Now count those that have at least two Bs. Then divide.

Let X = Number of boys from three children, then you may treat X as being binomially distributed. That is: $$X\sim Bin(~n=3~,~p=\frac{1}{2}~)$$ Where I have assumed that the probability of having a boy is equal to the probability of having a girl.

So what you are trying to find is:

$$P(X\ge2 ~|~X\ge1)=\frac{P(X\ge2 ~\cap~X\ge1)}{P(X\ge1)}=\frac{P(X\ge2 )}{P(X\ge1)}$$ Where I have used Bayes Rule and the fact that the probability of having at least 2 boys AND having at least 1 boy is the same as just having at least 2 boys.

Evaluating the numerator: $$P(X\ge2)=P(X=2)+P(X=3)=\binom{3}{2}\left(\frac{1}{2}\right)^3+\binom{3}{3} \left( \frac{1}{2}\right)^3=\frac{1}{2}$$ I'll leave the rest to you

• this gives 4/7 same as my intuition Sep 20 '14 at 5:44

Assuming that the probability of having a child being male/female is $\frac{1}{2}$, then you can simply apply a tree diagram.

Applying our knowledge of conditional probability (or just reading off the tree diagram), it should be $\frac{4}{7}$

but the solution given in book goes as --->

Event = having at least two boy

The event is occurring under the following situations:

• second is a boy and third is a girl OR
• second is a girl and third is a boy OR
• second is a boy and third is a boy

so the probability will be (1/2) *(1/2) + (1/2) *(1/2) + (1/2) *(1/2) = 3/4

That's the solution to the problem of finding the probability that at least two children are boys when given that the first child is a boy.

$$\underbrace{\overbrace{BBB,BBG,BGB}^{\text{these }3},BGG}_{\text{of these }4},GBB,GBG,GGB,GGG$$

If you are only given that at least one child is a boy, then that might be the first, second, or third child.

$$\underbrace{\overbrace{BBB,BBG,BGB,GBB}^{\text{these }4},BGG,GBG,GGB}_{\text{of these }7},GGG$$