Why my answer for this conditional probability problem is wrong?

Question

A family has two children. What is the probability that both the children are boys given that at least on of them is a boy?

Solution given in my book is

My doubts and my solution

If a family has a child and you are required to find that what is the probability of it to be a boy then your answer will be 1/2.

Now in the present problem it is given that out of two children one is a boy then what is the probability that both are boys.

One child is boy the second child can be a boy or a girl, chance of boy is 1/2.

$P(boy|boy)=\frac{1}{2}$

What I mean is that there are two possible outcomes after we have a boy, it can be a boy or a girl out of this boy is a favorable outcome and so we have 1/2.

Request

Please don't mark this problem as duplicate of this problem or of this problem because my logics are different.

Answer 1

You would be correct if $F$ was "the eldest child was a boy", or such information which gives order.   Then you could evaluate the probability that the other child is also a boy the way you suggest.

However the actual event is "at least one child is a boy" and that is not the same thing.   That does not specify which of the two children might be a boy; only that both are not girls.   There's one way both could be boys, and two ways one could be a girl, and these three ways are equally probable so...

What I mean is that there are two possible outcomes after we have a boy, it can be a boy or a girl out of this boy is a favorable outcome and so we have 1/2.

There are two possible outcomes, but they are not equally likely.

Answer 2

Always keep in mind that when you solve for a conditional probability, make sure you are using $\frac{P(\text{successful})}{P(\text{total})}.$ You're total ways here is NOT $2$, since there are actually $3$ ways to get at least one boy: BB, BG, and GB. Only one of these is "successful," namely, BB. Since each of the total $3$ ways is equally likely, the correct probability is $\boxed{\frac{1}{3}}.$

Answer 3

Ever hear the puzzle "I have two coins; they add up to 30 cents and one of them is not a quarter; what two coins are they". The answer? One is a a nickle and the other is a quarter. Wait? You said one of them was not a quarter! It isn't; it's a nickel and the other one is a quarter.

Believe it or not that is legitimate and useful.

You know one of them is a boy but you don't know which one. There are four equally likely possibilities. One is a boy and the other is a boy. One is a boy and the other is a girl. One is a girl and the other is a boy. One is a girl and the other is a girl.

The last one is ruled up. So we are left with 3 equally likely situations. One is a boy and the other is a boy. One is a boy and the other is a girl. One is a girl and the other is a boy.

One situation is 2 boys. Two situations are 1 girl and 1 boy.

So the probability of 2 boys is 1/3. The probability of 1 boy and 1 girl is 2/3.

The other way to view it is to arbitrarily think about which one is older. This is equivalent to thinking of two dice as being different colors to show that the dice rolls are independent. This is the exact same thing. The childrens' genders are independent.

Suppose you roll two dice. One green and the other is red. You are told one of them rolled even but you aren't told which one. What is the probability both rolled even?

So the guy says "I have two children. The oldest in name Qinkleberrysmith and has green skin. The youngest is name Schmithbuttflangy and has red skin. Oh, one of them is a boy."

So what is the probability that they are both boys? Well, Qinkleberrysmith could be a girl while Schmithbuttflangy is the boy. Or Schmithbuttflangy could be a girl and the Qinkleberrysmith could be the boy. Or they could both be boys.

Now does it actually matter if we know that one is older than the other or what their names are or what the color of their skin is? No, it does not.