P(B) = \(\frac{3}{5}\)and P(A’ ∩ B) = \(\frac{1}{2}\) are given.
∴ P(A’ ∩ B) = P(B) – P(A ∩ B)
∴ \(\frac{1}{2} = \frac{3}{5}\) – P(A ∩ B)
∴ P(A ∩ B) = \(\frac{3}{5} – \frac{1}{2} = \frac{6−5}{10 }= \frac{1}{10}\)
Now, P(A|B) = \(\frac{P(A∩B)}{P(B)}\)
= \(\frac{1}{10}\frac{3}{5} = \frac{1}{10}×\frac{5}{3} = \frac{1}{6}\)
∴ P(A|B) = \(\frac{1}{6}\)