S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6
Let A : Event of getting an odd number
B : Event of getting a prime number
A = {1, 3, 5} ⇒ n(A) = 3
B = {2, 3, 5} ⇒ n(B) = 3
A ∩ B = {3, 5} ⇒ n(A ∩ B) = 2
∴ P(A) = \(\frac{n(A)}{n(S)} =\frac{3}{6} =\frac{1}{2}\)
P(B) = \(\frac{n(B)}{n(S)} =\frac{3}{6} =\frac{1}{2}\)
P(A ∩ B) =\(\frac{n(A\,\cap\,B)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)
∴ P(B/A) = \(\frac {P(A\,\cap\,B)}{P(A)}=\frac{\frac{1}{3}}{\frac{1}{2}}= \frac{2}{3}\) .
