1. ## Probabikity problem

Case (1) Suppose I meet you on the street corner, I introduce you to my son and inform you that I have another child. The probability that my other child is a girl is 2/3

Case (2) Now suppose I meet you on the street corner, I introduce you to my 1st born (or 2nd born) son and inform you that I have another child. The probability that my other child is a girl is 1/2

Now suppose in case 2, because of the noise in the streets, you did not hear me state that this son was my 1st born.

This is where I get confused. Whether I had said 1st born or 2nd born the probability that the other child is a girl is 1/2.
However, if you did not hear me say whether this son is my 1st or 2nd born child, then this is exactly like case 1 and the p(other child is a girl)=2/3.

What am I missing here???!!!

2. ## Re: Probabikity problem

Originally Posted by JaguarXJS
Case (1) Suppose I meet you on the street corner, I introduce you to my son and inform you that I have another child. The probability that my other child is a girl is 2/3

Case (2) Now suppose I meet you on the street corner, I introduce you to my 1st born (or 2nd born) son and inform you that I have another child. The probability that my other child is a girl is 1/2

Now suppose in case 2, because of the noise in the streets, you did not hear me state that this son was my 1st born.

This is where I get confused. Whether I had said 1st born or 2nd born the probability that the other child is a girl is 1/2.
However, if you did not hear me say whether this son is my 1st or 2nd born child, then this is exactly like case 1 and the p(other child is a girl)=2/3.

What am I missing here???!!!
is Case (1) given as part of the problem? It's certainly not true otherwise.

3. ## Re: Probabikity problem

Originally Posted by romsek
is Case (1) given as part of the problem? It's certainly not true otherwise.
Case1 and 2 were separate problems. Then I started to think what if I did not hear everything clearly--that is I did not hear which position the son was born in.

4. ## Re: Probabikity problem

Originally Posted by JaguarXJS
Case1 and 2 were separate problems. Then I started to think what if I did not hear everything clearly--that is I did not hear which position the son was born in.
you've lost me

how is Case (1) a problem? There's no question. Just a dubious statement about the probability of the guys other kid being a girl. W/o any other information this probability will be 1/2, not 2/3.

5. ## Re: Probabikity problem

Originally Posted by romsek
you've lost me

how is Case (1) a problem? There's no question. Just a dubious statement about the probability of the guys other kid being a girl. W/o any other information this probability will be 1/2, not 2/3.
OK, I will state case 1 as a problem.
Suppose I meet you on the street corner, I introduce you to my son and inform you that I have another child. WHAT is the probability that my other child is a girl?
Here is my result. The 3 cases are bb, bg and gb. Each of these 3 have the same probability and 2 of the 3 have a girl, so I conclude that p(my other child is a girl)=3/2

6. ## Re: Probabikity problem

Originally Posted by JaguarXJS
OK, I will state case 1 as a problem.
Suppose I meet you on the street corner, I introduce you to my son and inform you that I have another child. WHAT is the probability that my other child is a girl?
Here is my result. The 3 cases are bb, bg and gb. Each of these 3 have the same probability and 2 of the 3 have a girl, so I conclude that p(my other child is a girl)=3/2
You've introduced the order of birth where it doesn't matter. In the context of the problem bg and gb are the same.

It boils down to the possibilities of the other child, either b or g and without additional information, such as p[b|b] or p[g|b], the probability of both is 1/2.

7. ## Re: Probabikity problem

Originally Posted by romsek
You've introduced the order of birth where it doesn't matter. In the context of the problem bg and gb are the same.

It boils down to the possibilities of the other child, either b or g and without additional information, such as p[b|b] or p[g|b], the probability of both is 1/2.
I am missing something here. Forget for a moment that you know that I have a son. If I have 2 children then the sample space is {bb, bg, gb, gg}, xy means the sex of the 1st child is x and the sex of the 2nd child is y. Now I think that you are saying that bg and gb are the same. So then the probability of having a girl is 2/3 instead of 3/4. I am fairly certain that 2/3 is wrong as bg and gb are in fact different. I guess what I am asking is why in my case are bg and gb the same.
Thank you very much for the time you have spent with me so far.

8. ## Re: Probabikity problem

Originally Posted by JaguarXJS
Case (1) Suppose I meet you on the street corner, I introduce you to my son and inform you that I have another child. The probability that my other child is a girl is 2/3

Case (2) Now suppose I meet you on the street corner, I introduce you to my 1st born (or 2nd born) son and inform you that I have another child. The probability that my other child is a girl is 1/2!
Suppose that I tell that I have two children.(which is in fact true).
What are the possible outcome space? Is it:
$\begin{array}{*{20}{c}} {BB} \\ {BG} \\ {GB} \\ {GG} \end{array}$

In that list three of the pairs have at least one boy. So case I has a probability of $\dfrac{3}{4}~.$

In that list two of the pairs have a boy as the first. So case II has a probability of $\dfrac{2}{4}=\dfrac{1}{2}~.$
So knowing that the first-born of a pair is a boy means that the probability the second-born is a girl is $\dfrac{1}{2}$

9. ## Re: Probability problem

Originally Posted by Plato
Suppose that I tell that I have two children.(which is in fact true).
What are the possible outcome space? Is it:
$\begin{array}{*{20}{c}} {BB} \\ {BG} \\ {GB} \\ {GG} \end{array}$

In that list three of the pairs have at least one boy. So case I has a probability of $\dfrac{3}{4}~.$ Yes, there are 3 cases that has a girl, but one is GG which makes no sense as we KNOW that this person has a son in case 1. I think that you did not read case 1 clearly

In that list two of the pairs have a boy as the first. So case II has a probability of $\dfrac{2}{4}=\dfrac{1}{2}~.$
So knowing that the first-born of a pair is a boy means that the probability the second-born is a girl is $\dfrac{1}{2}$ This is true. Even if you knew that the 2nd born is a boy, then the probability the second-born is a girl is $\dfrac{1}{2}$ as well