Let B= boy G= girl
And let us consider, in a sample space, the first child is elder and second child is younger.
Total possible outcome = {BB, BG, GB, GG} = 4
Let A = be the event that both the children are girls = 1
Therefore P(A) = \(\cfrac14\)
Case 1.
Let B = event that youngest is girl = {BG, GG} =2
{Since we have considered second is younger in a sample space}
Therefore P(B) = \(\cfrac24\)
And (A ∩ B) = both are girls and younger is also girl = (GG) = 1
Therefore , P (A ∩ B) = \(\cfrac14\)
We require P(\(\cfrac{A}{B}\))
\(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\)
= \(\cfrac{\frac14}{\frac24}=\cfrac12\)
Case 2.
Let B = event that at least one is girl = {BG,GB GG} =3
{Since we have considered second is younger in a sample space}
Therefore P(B) = \(\cfrac34\)
And (A ∩ B) = both are girls and atlas one is girl = (GG) = 1
Therefore , P (A ∩ B) = \(\cfrac14\)
We require P(\(\cfrac{A}B\))
\(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\)
= \(\cfrac{\frac14}{\frac34}\) = \(\cfrac13\)