probablility again?

Question

Suppose that the probability of a male birth is 0.5. A couple wishes to have exactly two female children in their family. they will have children until this condition is fulfilled.

question

what is the probability that the family has 4 children??

Answer 1

You're probably going to have to look at different probabilities to get this one. It is not the Poisson Distribution, since a decision is based on an prior outcome (Bayesian Prob), of which I draw a complete blank. Drawing a probability tree is an approach, too

P(2 girls in a family with 2 kids)= 0.25
Note only 75% of families "advance to the next stage, having 3 kids.
2/3 of this 75% of total have 1 boy and 1 girl, so a 50% girl outcome means 1/2 of these can "drop out" at 3 kids. Of the other 1/3, which have 2 boys, all must "advance" to a 4-child family.

At this point P(2 girls/3 kids) = .25
The probability that a family has 2 or 3 kids with 2girls is 0.5

Moving on to the next "round", there is a .125 probability that there are 3 boy families, and regardless of what happens with a 4th child, these families will have to have more than 4 children if they are to have 2 girls. There is a .125 prob that the family has 2 boys and then a girl. So if the 4th child is a 50/50 event, 0.0625 prob that they will be BBGG and can quit. There is also a .25 probability that they had a boy and a girl (order immaterial) and then a boy. Half of these families are expected to pick up a second girl for the 4th kids, or a probability of 0.125.

So the probability that a family with 4 kids has 2 girls, having decided to "quit" having kids when they have their second girl, is 0.1875

Answer 2

The only branches of length 4 are:
gbbg,
bbgg,
bgbg,
therefore the probability is 3/16.

Answer 3

I'll give you a hint. It's a negative binomial distribution. Let me know if you need more hints.

And yes, the answer is 3/16 = 0.1875