(i) P(E∪F) = P(E) + P(F) – P(E∩F)
= 0.68 = 0.6 + 0.2 – P(E∩F)
= P(E∩F)=O.12 P(E) x P(F)
= 0.2 x 0.6 = 0.12 = P(E∩F)
Hence E and F are independent events. (ii) (a) Let X denotes the random variable of number of odd number in the throw of a die 6 times. Clearly X has a Binomial Distribution with n =6 and \(p=\frac{3}{6}=\frac{1}{2}\)