Bizarre Probability Problem (1 Viewer)

[SystemShock]

That’s a strange analysis and I’m fairly sure it is wrong. The only thing that matters is whether you know there is only 1 girl or whether the question reveals a clue of how many girls there are. Chuck made a good case that the question doesn’t give that away due to word play. That is the only basis for anything other than a 50% chance the second child is a girl. You’re adding qualifiers/characteristic of birth order, which I stated does change things. If the question was what are the odds of having 2 girls given the first born was a girl, then I would use the matrix of 3 possibilities (bb, bg, gg), and the answer would be 1/3rd. If you don’t know they have 2 girls, then the odds of the next child being a girl is 50%, no matter what name the other child has.

You have the same issue with the wording of the problem. I thought you mentioned you were an engineer?

I code, so I have the same issue with adding/assuming certain information. And you can't do that in coding. Unless more variables are declared, I can only look at the one that that's properly declared and go for the simplest, most obvious answer.

 

[UncleTrvlingJim]

I think this problem is supposed to say in the second set that the older child is a girl...

And then the answers are 1/3 and 1/2 respectively. It's a pretty well known probability problem in math circles.

 

[zeetes]

Where is @zeetes he has a big brain right?

unfortunately big brain is mostly air.

 

[FullMonte]

I think this problem is supposed to say in the second set that the older child is a girl...

And then the answers are 1/3 and 1/2 respectively. It's a pretty well known probability problem in math circles.

No, the second set is supposed to say the child's name, or hair color, or eye color, or whatever. If you know a detail about the child that occurs more often than once, the probability that the other child is also female becomes 1/2.

Look at it this way. Assume there are 1000 pairs of children, and the sexes are equally distributed. That means that you'll have 250 BB pairs, 250 BG pairs, 250 GB pairs, and 250 GG pairs. Knowing that one child is a girl eliminates the BB pairs, which leaves us with 500 pairs that contain one of each sex, and 250 pairs that are both girls. Now, we learn a detail about the child, such as her name. We can assume that in a group of 1000 girls, that name will occur a certain number of times (for argument's sake, we'll say that the name Debra occurs once in every 10 girls). That means that in the 500 pairs that contain one of each sex, there are 50 girls named Debra. In the 250 pairs that are both girls, there are also 50 girls named Debra. So, if the girl is named Debra, there is an equal chance that she has a brother or a sister. The reasoning is that while there are two times as many pairs that are BG/GB pairs, there are twice as many girls in the GG pairs, so the chances equal out.