Question

Please anyone can help me with this probability question and please provide explanation. Thank you so much!

1. (i) Consider two events A and B, with P(A) = 0.3, P(B) = 1. Compute P(A∩B), P(Ac ∩B), and P(A ∩ Bc) (where we denote by Ac = Ω \ A the complement of an event A).

   (ii) We now consider three independent events A, B, and C. Using the definition of inde- pendence, show that the two events Ac and Bc ∪ Cc are independent.

Answer 1

ACAB Sch. W Given P(A) = 0.3 , P(B) = 1 Clarly ACB So AnB = A @ P(ANB) = P(A) = 0.3 And OPAB) = P(B) – PCA) = 1 - 0.3 = 0.7 And. © P(ANB) -o E AnB = 6] (W A, B and C are independent, so P(ANB) = P(A). PCB) PANCI = P(A). (c) PCBNC) = PCC) AC ,PANCAB) = P(A).MB)• PCC) Consider Pan (Buco) =P[ n(Bncje ]using De Morgan's Law ? (Bncje = Bruce = p [AUCBnc}}] fusing De Morgan's Law..] (AUBJC = AAB = 1- P[A U (BNC) =1- (P (A) + P (BNC) - P(ANBAC) -1- PA) – P(B). PIC) + PCA) - P(G)- PC) =(1-PCA)] – PIB).fc) [1 – P(A)] = (paj] (i - P(B). Ples] = (1-P(A)] . (T-P (BNC)] = P(AC). [Back] = P(AC) . P (BUC) [uring De Morgand law] so AC and Bruc are independent