Question

A family has two children. What is the probability that both the children are girls given that at least one of them is a girl?

Answer 1

A family has two children.

∴ Sample space S = {BB, BG, GB, GG}

∴ n(S) = 4

Let event A: At least one of the children is a girl.

∴ A = {GG, GB, BG}

∴ n(A) = 3

∴ P(A) = \(\frac{n(A)}{n(S)} = \frac 14\)

Let event B: Both children are girls.

∴ B = {GG}

∴ n(B) = 1

\(\therefore P(B) = \frac{n(B)}{n(S)} =\frac 14\)

Also, A ∩ B = B

∴ P(A ∩ B) = P(B) = \(\frac 14\)

∴ Required probability = \(P\left(\frac BA\right)\)

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

\( = \cfrac{\frac 14}{\frac 34}\)

\(= \frac 13\)

Answer 2

Let B denote a boy and G denote a girl. 

Then the sample, S = {BG, GB, BB, GG}. 

∴ n(S) = 4 

Let E be the event that both children are girls. 

Let F be the event that atleast one of them is a girl. 

Then E = {GG}, n(E) = 1 

F = {BG, GB, GG}, n(F) = 3 

P(F) = \(\frac{n(F)}{n(S)}=\frac{3}{4}\)

E ∩ F = {GG}, n(E ∩ F) = 1 

P(E ∩ F) = \(\frac{n(E∩F)}{n(S)}=\frac{1}{4}\)

Required Probability P(F/E) = \(\frac{P(E∩F)}{P(F)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\)