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