A sample of four families with three children each was taken.find the probability that one family has two girls,two families have three girls and one family has three boys.

Answer 1

Pr(three boys,1g)=1/16
pr(2g,2b)=1/16
pr(3g,1b)=1/16

Pr(above)=4*pr(3b)*3pr(3g)*2pr(3g)*pr(2g)=4*1/16*3*1/16*2*1/16*1/16
=4! (1/16)^4
check that.

Answer 2

I disagree
The probs that Bob found would each have 4 children, not three

prob(of 3 children, two are girls)
= C(3,2)(1/2)^2 (1/2) = 3/8

illustration:
BBB
BBG
BGB
GBB
GGG
GGB -- 2 GIRLS
GBG -- 2 GIRLS
BGG -- 2 GIRLS
So there are 3 cases of 2 girls
prob(2 girls) = 3/8

prob(3 girls)
= 1/8

prob(3 boys)
= 1/8

so prob(as stated in problem)
= (3/8)(1/8)(1/8)(1/8)
= 3/4096