Question

Challenge Problem: Suppose that we toss a biased coin repeatedly with p being the probability of heads and q = 1 - p being the probability of tails. In this exercise we will compute the probability that there is a run of m heads in a row before there is a run of s tails in a row and express the probability in terms of p, q, m and n. (a) (1pt) Verify the following two identities. (You only need to verify one of them, the other one is true for the same reason) P(A|BC) = P(A|Bºn C)P(C|Bº) + P(A|Bºn Cº)P(CC|BC) P(AB) = P(ABC)P(C|B) + P(AB C')P(CC|B) (b) (1pt) Now let A = {run of m heads precedes a run of n tails}, B = {1st toss is heads} and C = {first n tosses are tails}. Use the identities from part (a) to show: P(A|B“) = P(A|B)(1 – q-1) P(A|B) = pm-1 +P(A|B)(1 – pm-1) (c) (1pt) Solve for P(AB) and P(A Bº) in part (b) and use the first Bayes theorem to obtain pm-'(1 – q") P(A) pm-1+qn-1-pm-'qn-1

question B needs more explaination

Answer 1

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a To prove P PGALBC) = PC ALB806). PC C) BC) te(Albene). P[<AB), R.H.S. P(A/B nc). P(cl be + P(Alb996.). P(c)B) - PLEAB Plan Binc) Pas P(B) + Plano enco) PCCB Pored (6) f(BC) plano me) + PC Anbeno Elf (Anenc) " (A18ncy} [ Anbnc & AnBnce PCB are mutually exclusive

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Similarly. P[c418)= P[{i". n tosses are not all fails?][Cist ails} 1-PC18 laq (2

Now , le its given that PLAlbince) so its given Bence Ist toss is head also that ist n- tosses are not all tails. This mean at some point there'll be a heads to break the sequence (so that not all first n-tosses are tails). As it A occurs then C has to occur A c cc So only the information from that at one I toss there will bea head in the 1 st - tosses: And from there we start fresh as A given. heads PC AL BENC) =PT (previous toss is) =PA1B) the experiment is rebooted from the tosses where heads' come for the first time between 1st toss to nt toss Ć given is n- Lone toss is

from O se we have, T Therefore, PCAIB) 8°) =P PC Albinc).P 2 + ?(416). (1-21)- ho nc} = پیالہ ist tails toss is beats and all S+ netesses are tails z all first na tosses are tails. C Benc Bcnc = c ! There tosses have alrea nun Tina as CCB already Completed their oy na tails row. " A can't occur now ③ Hence of ?(.A. | BERC) = P(ALC) using we get, PLA|89) = 0 + PCA)B). (L=9**) 3 in.

Now its given that B has occured so PLA1B). Now ß given U occur either in the or later. first tosses itself P(there a son of m-) heads list forss P(ALB) Z is heads list toss occurs later as the son of mal heads did not occur istieads

mal + PLA A occurs occurs yet P is after (m-1) run of heads was cut by to heats) a tails ma Serena theads t m') PLAB (as as again after a Tails cut the sun of the experiment reboots' with only the information that previous toss was tails Hence the prob. that run of heads gets before (1-pm-1 (: as 1st is & system reboots' with information of where the first Tails. of JP (ALB A185) cot m-muns already heads cut this run heads. ... PC A1B) => t PLATE_PCA |B)P(69) + f(A169) PCE). (using 1st Bay es. Theorem).

the equations. in part Now, Solving: we get, P(A/B) ابيا Tot p ^^)(1-4) m - L 5 ) P(ALS) 1-2 PLAIB 1-(-2)--) E P(4189) =_**! (1-2-2 1-(1-2-1){1-p4-1) Thenfore, m-1 P[a) & p... 1- (1913(1-bmcy) + (1-P)*__b1/ m -(1m9 -1) (1-3-1-1

1 - щ & PLATE (+ (1-6) (1991) n-1 1) (1-bem m-1 n-1 nal U t -q - tpqh m-1 dt +qn-1 qu-1pm- -l be 1-1) b m-1 n-/ q u PlA) 41-9") tqh-1-pm-light [- q-1-p pm-1 tq