Given: P(A) = 0.3, P(B) = 0.6.

(i) P(A and B)

As A and B are independent events.

⇒ P(A and B) = P (A ∩ B) = P(A) . P(B)

= 0.3 × 0.6

= 0.18

(ii) P(A and not B)

⇒ P(A and not B) = P (A ∩ B^’) = P(A) - P(A ∩ B)

= 0.3 - 0.18

= 0.12

(iii) P(A or B)

⇒ P(A or B) = P(A ∪ B)

As we know, P (A ∪ B) = P(A) + P(B) - P (A ∩ B)

⇒ P (A ∪ B) = 0.3 + 0.6 – 0.18

⇒ P (A ∪ B) = 0.72

(iv) P(neither A nor B)

P(neither A nor B) = P(A^’ ∩ B^’)

As, { A^’ ∩ B^’ =(A ∪ B)^’}

⇒ P(neither A nor B) = P ((A ∪ B)^’)

= 1 - P (A ∪ B)

= 1 - 0.72

= 0.28