Let A: Getting two heads

B: At least one coin showing a head.

**S = {HH, HT, TH, HH} **

Then, A = {HH}, B = {HT, TH, HH} ⇒ A ∩ B = {HH}

∴ P(A) = \(\frac{n(A)}{n(S)} =\frac{1}{4}\) , P(B) = \(\frac{n(B)}{n(S)} =\frac{3}{4}\),\(P(A\,\cap\,B) = \frac{n(A\,\cap\,B)}{n(S)} = \frac{1}{4}\)

### Now, Required probability = P(A/B) = \(\frac{P(A\,\cap\,B)}{P(B)} = \frac {\frac{1}{4}}{\frac{3}{4}}\) = **\(\frac{1}{3}\)**