2
$\begingroup$

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.

EDIT* -->

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

$\endgroup$
4
  • 1
    $\begingroup$ Are all scenarios equally likely? $\endgroup$ – copper.hat Sep 20 '14 at 4:40
  • $\begingroup$ @copper.hat yes $\endgroup$ – romitheguru Sep 20 '14 at 4:52
  • $\begingroup$ If the question is stated the way you have asked it, then the answer is 4/7 $\endgroup$ – WeakLearner Sep 20 '14 at 5:35
  • $\begingroup$ 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 $\endgroup$ – WeakLearner Sep 20 '14 at 5:37
0
$\begingroup$

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.

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ this gives 4/7 same as my intuition $\endgroup$ – romitheguru Sep 20 '14 at 5:44
0
$\begingroup$

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 $\frac{4}{7}$

$\endgroup$
2

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.