Question

(a) Let P(B1∩B2)>0, and A1∪A2⊂B1∩B2. Then show that

P(A1|B1).P(A2|B2)=P(A1|B2).P(A2|B1).

(b) Let A and B1 be independent; similarly, let A and B2 be independent. Show that in this case, A and B1∪B2 are independent if and only if A and B1∩B2 are independent.

(c) Given P(A) = 0.42,P(B) = 0.25, and P(A∩B) = 0.17, find

(i)P(A∪B) ;

(ii)P(A∩Bc) ;

(iii)P(Ac∩Bc) ;

(iv)P(Ac|Bc).

Answer 1

Complete solution is given in attached images:

а) te have (An U.AZ) C (Ben B27 А, СВ, ОВ, д А. СВ.,ПВ, А, С, А, СВ, А, СВ, А, СВ. А, с А, ОВ, от А = A, B, к А, = A, Пр, от А, A, B, (2) We -Киод (А. ) . f (А.) : P(A) P(A) Ffrom - р (А, ов) - (А, В,) = e (A, B,) - Р (А. Лв.) 3 Р(А, ПВ) P(A₂0 B₂) КА, Пв, P(A₂02B) P (B) P (B2) P_BƏ. P(B) Э PA, IB). P (A2/B2 Р (А, В,) : Y (А, ІВ) Hence prove.

Now A & B Lis independent 3)_P(ANB) = f(A). P (B1) A & B2 is independent = P(ANB) P(A). P (B) if a - Now A & B.B. are independent then Plan B,16,2) = R(A): PCB.0 Ba? p (An (6.1.6.)) = PLANBA) 0 (ANB) = P(ANB) + P(ANB) - plane, n Anex) P(A) PlB.) + Pla) P (82) - Plan (Bin B2) = PA Plbi) + P(A) (8.) - PP (B.NB) = P(a) P@) +P (B) – p (2.0.6 : PA. PB, B2) A & B, RB indenpendent then A & B, UB₂ are_independent. Hence if are then Now if A & B, DB2 indenpendent P An (B, B2) PIA). PB. 1B22 plan@)!) (ANB) = P(A): P (B, VB.) > P An 6) + Plan Be) - PAn (8,16.) = P(A). PB, UB,) b pla). P(B) + PCA).PB) - plan e, ne.) = P(A). PB, UB2)

- l(A).P (80) + P(A): P[82-PA):P (B,18,)=_ pan P (AN (8,082) 3 pca) [p@.)+ P(6.) - Rebo PCB,18.) = plan (8,08>) = P(A) P (B, NB) plan (8,982) Hence if are Af Bill Bz A & B, 0 B independent independent. A (IT) then are fi II لبا if and only from proof 6 get, and B, 1 B 2 are independent if . A and BrO Belare independent. Hence proved. 4. e] hiven PA) : 0.42 = 0.42, Pb= 0.25, P(ANB) = 0.17 i Plaub) = P(A) + P(B)- P(ANB) = 0.42 to.25+ 0.17 o.so

e) ii) P (ANB) P(A) - P(ANB) 0.42-0.17 11 0-25 iii) PARB) 1- P(AUB) 1-0.5 0.5 iy) P (A/B) P(AnB? PB9 PA AB) 1- PB 0.50 0.75 0.6666

Thank You.