Easy(?) Probability

I posted a similar problem on here a couple of years ago, if you
remember that don't contribute, just sit back and enjoy the discussions!

I phone a work colleague up and his daughter answers the phone. He has
previously told me that he has two children at home, I had no idea he had
a daughter. What are the chances that the other child is also a girl?

Originally posted by wolfgang59
I posted a similar problem on here a couple of years ago, if you
remember that don't contribute, just sit back and enjoy the discussions!

I phone a work colleague up and his daughter answers the phone. He has
previously told me that he has two children at home, I had no idea he had
a daughter. What are the chances that the other child is also a girl?

Agreed =)

Well the items are (elder sibling/younger sibling):

b/b (evidently untrue)
g/b
b/g
g/g

so 1 in three should be correct.

b/b (b1 answers) nope
b/b (b2 answers) nope
b/g (b answers) nope
b/g (g answers)
g/b (g answers)
g/b (b answers) nope
g/g (g1 answers)
g/g (g2 answers)

I'm going to say 1/2

Why would it be any different than:
"I flip a coin, and it comes up heads, what are the chances that my next flip is heads?"

You could use the same logic that you guys just used if you had already flipped the coins and were asking after the fact.

Note that this is different than the problem:

I have four socks, two black and two blue. I randomly put to of the socks in a bag.

You grab one sock out of the bag, and it is black. What is the probability that the other sock is black?

So... before the phone call the probability that both of the children were girls was 1/4.

After the phone call the probability that both of the children are girls is 1/3 and the probability the the other child is a girl is 1/2.

Is this a correct line of reasoning, or not?

Originally posted by joe shmo
So... before the phone call the probability that both of the children were girls was 1/4.

After the phone call the probability that both of the children are girls is 1/3 and the probability the the other child is a girl is 1/2.

Is this a correct line of reasoning, or not?

I know that if this person has two children, one of the a girl called sue then the probability the other one is a girl is 1/3.

If a person has two children, the oldest one is a girl called sue then the probability is now 1/2.

I would go with the BB, BG, GB, GG explanation and go with 1/3 however often the oldest child is the one who answers the phone.

Let's take an infinite set of families with two children, where children have an equal probability of being boys or girls, and you ask the question:

"Of all of the families with at least one girl, what is the probability that both children in that family are girls?"

Then the answer is 1/3

If, on the other hand, you randomly choose one family, and then randomly sample one of the children, and discover that child is a girl, then the probability that the other child is a girl is not conditionally related to your sample and the probability is 1/2.

KEEP GOING GUYS 😀

Of course if you phone the work colleague up somewhere that is not his home it could be one of his 5 children that have left home.

If you have four two children families BB, BG, GB, GG. You take a random sample of one by for instance phone. There is only one case in which the other child matches your sample (and one case which is clearly excluded by the result of the sample). So I am still sticking with 1/3.

Originally posted by deriver69
Of course if you phone the work colleague up somewhere that is not his home it could be one of his 5 children that have left home.

If you have four two children families BB, BG, GB, GG. You take a random sample of one by for instance phone. There is only one case in which the other child matches your sample (and one case which is clearly excluded by the result of the sample). So I am still sticking with 1/3.

I thought again.. yep. 50%

50/52


GRANNY.