MATHEMATICAL TRIPOS: PART IA PROBABILITY Example Sheet 2 (of 4) Lent 2023 JRN 1. A coin with probability p € [0, 1] of heads is tossed n times. Let E be the event 'a head is obtained on the first toss' and Fx the event 'exactly k heads are obtained'. For which pairs of non-negative integers (n, k) are E and FÅ independent? P(E) = P P(FK)= (1) pk (1-p) ntc n-k P(EnFW)= P.(K-1)plc-1(1-P) if independent, [AFK) = (E). PIFK ). n-je P. (K-1)pkt(1-P) =8-12)pk (1-pjak (h-1) !: P. n! K! In+51 (-1)! (n + ) ? 1 = p. n K K - = P P(F)-P(FC) = p. 11-p)n. $ 0 = PLENFic) 16=0 PIEAFK) = 0. not independent. n=0. P(EnFK) = 0. P(E) . P(F) = 0.0 = 0. E and Fic is independent for =P, PETOi], K?, I,n#01 and { n=o. KEIR }.
2. The events A and B are independent. Show that the events Ac and B are independent, and that the events Ac and Be are independent. AUB = 2 AVAC=2 130B=2 AUB = (AUB) ( (AVAC). P (ANB ) = P(A ) . P(B ). P(A) = 1-P(AC) = AU (BRAC) (P(AVI3) = IP(A) + IP(BNAC). (PIBRAC)= P(AVB)-IP(A) " A and AC disjoint Vi. A and BRAC disjoint = (PCA) +P(B) -PLANB) -P(A) = (P(B) - IP(ANB). = IP(B)- IP(A)·(P(B) - IPIBY-ID(PS) + IP(AS IP(B) IP(A) = 1-IPCAC) = (P(AC) [(B) Therefore. B and AC are independent. Boy Symmetric, A and BC are also independent
A VI3 = (AUB) ( (A VAC) = A VCBNA9) A and AC disjoint A and (B" A ) dijon. (PCA V13) = IP(A) + IP(BNA) IPCA (BC) = IP(AVBC) - IPCA) = IP(A) + IP(BC) - (PCAn BC) - (PLA) E = [P(BC ) - IP(A)- PIBC ) = [(BC) - IP(BC ) + IP (AC)- P(BC) = IP(AC) ·(PCBC) Therefore, AC and BC are independent A and B independent