A coin has probability p of showing head when tossed. It is tossed n times. Let Pn denote the probability that no two (or more) consecutive heads occur. Prove that p1 = 1, p2 = 1 – p2 and pn = (1 – p). pn - 1 + p(1 – p) pn - 2 for all n ≥ 3.

7 years ago
Hello Student,
Given that the probability of showing head by a coin when tossed = p
∴ Prob. of coin showing a tail = 1 – p
Now pn = prob. that no two or more consecutive heads occur when tossed n times.
∴ p1 = prob. of getting one or more on no head = prob. of H or T = 1
Also p2 = prob. of getting one H or no H
= P (HT) + P (TH) + P (TT)
= p (1 – p) + p(1 – p) p + (1 – p) (1 – p)
= 1 – p2, For n ≥ 3
Pn = prob. that no two or more consecutive heads occur when tossed n times.
= p (last outcome is T) P (no two or more consecutive heads in (n – 1) throw) + P (last outcome H) P ((n – 1)th throw results in a T) P (no two or more consecutive heads in (n – 2) n throws)
= (1 – p) Pn -1 + p (1 – p) pn – 2 Hence Proved.

Thanks
poiuhgfds
15 Points
one year ago
Thank you
For n≥3, if the last outcome is T, then the probability that the first (n−1) tosses do not contain two (or more consecutive heads is p n−1 ​ and if the last outcome is H, then (n−1)th outcome must be T and the probability the first (n−2) tosses do not contain two (or more) consecutive heads is p n−2 ​ . Hence, p n ​ =p n−1 ​ ×P (nth toss results in a tail)+p n−2 ​ ×P (nth toss results in a head and (n−1)th toss results in a tail) =(1−p)p n−1 ​ +p(1−p)p n−2 ​ A) When n=1, then two possible outcomes viz. H and T satisfy the condition that two (or more consecutive heads do not occur. Thus, p 1 ​ =1. B) When n=2m the possible outcomes are HH,HT,TH and TT. Out of these, first outcome viz. HH does not satisfy the condition that no two (or more) consecutive heads occur. Thus, p 2 ​ =1−P(HH)=1−pp=1−p 2 . C) p 3 ​ =(1−p)p 2 ​ +p(1−p)p 1 ​ =(1−p)(1−p 2 )+p(1−p)1=1−2p 2 +p 3 D) p n ​ =(1−p)p n−1 ​ +p(1−p)p n−2 ​