Question

2. (a) Let P(Bin B2) > 0, and AUA, CBin B2. Then show that P(A/B).P (A2|B2) = P(A|B2).P (A2|Bi). (b) Let A and Bbe independent; similarly, let A and B, be independent. Show that in this case, A and B U B2 are independent if and only if A and Bin B2 are independent (c) Given P(A) = 0.42, P(B) = 0.25, and P(An B) = 0.17, find (i) P (AUB); (ii) P(An B°); (iii) P(A n B); (iv) P( ABC).

Answer 1

2-e, - al 2 (a) Since A and B, are independent, so PCAN B,) = P(A) PCB). > ☺ Since A and By are independent, so PCANB2) - P (A) PCB2) < (7) Let A and (B,082) be independent P[ An (B,032)] - PCA). PCB,0 B2) Now, p[ An (B, UB2)] P[AnB,) U PCAO B2)] PCA0B,) + PCAO B) - P[(ANB)(ANB2)] PCAOB,) + PCAOB2) - p[ An (B,0B2) 7 PCAO B) + P( A2) - P(A). PC,002) PCA). PCB) + PCA). P(82) - P(A)• PCB, n B2) P(A) [ P(B1) +PCB2) - PCB,OB2)] P(A) PCB,UB2) · 7.2.) P[an (B,UB2)] - P(A). PCB,UB2) This shows that Å and B, UB) are if A and B,9Bz) are assumed to be independent. Conrensely let A and B, UB2) be independent. So coe have, P[AO (BUB2)] PCA). PCB,0 B2) = - - A are independent

Now, - P[AO(BIO B2)] P[ (ANB) An B2)] PCANB, ) + PCAN B2) - P[ An B,) UCAN B2)] P(A) P(B1) + P(A) PCB2) - P[ ACB, UB2)] j[using) and ] PCA). PCB,) + PCA) P(B2) - PCA). PCB, UB2) [ey assumption] PCA) [ PCB,) + PCB) - PCB, UB2)] PCA). PCB, O B2) yhis cohows that A and Bin B2) are independent if it is assumed that A and (UB) independent % A and B U B2) are independent if and if and only if A and (BAB) are are independent. are we have given, PCA) = 0•42, PCB) = 0.25, PCAOB) = 0.17 PCAUB) = P(A) +PCB) - P(ANB) O•H2 +0.25 -0.17 0-67-0.17 = 0.50 (12) PCA ов°) = PCA) – PCAOB) = 0.42-0.17 -0.25 (iii) PCACOBC) - P[AUB) C) = 1 - P(AUB) P[(AUB) C) = 1 - P(AUB) = 1-0.50 - 0:59. (iv) P(AC BC) Plan B9) 0.50 PCAC%B9) PCB) 1-PCB) 1-0.25 0.75 0.50 dilig