Question

Answer the following with TRUE or FALSE, and justify your answer (this does not necessarily have to be a formal prooW). (a) If two events A, B where P(A) > 0, P(B) > 0, P(AN B) > 0 and A, B are independent, then AC and BC are independent as well. (b) For two events A, B where P(A) > 0, P(B) > 0, they can be disjoint and independent at the same time. (c) For two events A1, A2 where P(Ai) > 0 for i = 1,2 and A1, A2 independent, P(AJ U A2) = 1 – P(Af)P(A2). (d) For two disjoint events A and B where P(A) > 0, P(B) > 0, P(A|B) = P(BA).

Answer 1

True Since A and B are independent Rightarrow p(A intersection B) = p (A) .p (B) is A and B are independent we can show p (A intersection B) = p (A) p(B) proof p (A intersection B) = p (A union B) = 1 - p (A union B) = 1 - (p (A) + p(B) - p (A intersection B)) = 1-p(A) - p (B) + p (A) p(B) (By (1)) = p (A) - p (B) .p (A) = p(A) .p(B) (b) false for eg S = {1, 2, 3, 4, 5, 6} E_1 = {1, 3, 5} E_2 = {2, 4, 6} p(E_1) = p(E_2) = 1/2 p (E_1 intersection E_2) = 0 notequalto p (E_1) .p (E_2) Therefore E_1 and E_2 are not independent even though they are disjoint (c) True Since A_1 and A_2 are independent Rightarrow p (A_1 intersection A_2) = p (A_1) p (A-2) - (1) p(A_1 Union A_2) = 1 - p (A_1 intersection A_2) = (De - morgan�s law) = 1 2 - p (A_1) p (A_2) (By (1))