A family has 2 children, then Sample space = S = {BB, BG,GB,GG}, where B stands for Boy and G for Girl. 

(i) Let A and B be two event such that 

A = Both are girls = {GG} 

B = the youngest is a girl = {BG, GG}

Now, P(A/B) = (P(A ∩ B))/P(B) = (1/4)/(2/4) = 1/2

(ii) Let C be event such that 

C = at least one is a girl = {BG,GB,GG}

Now, P(A/C) = (P(A ∩ C))/P(C) = (1/4)/(3/4) = 1/3

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(\(\cfrac{A}B\))

\(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\)

= \(\cfrac{\frac14}{\frac34}\) = \(\cfrac13\)