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

Question

Aditi Chauhan · Answer

Hello Student,

Please find the answer to your question

Given that the probability of showing head by a coin when tossed = p

∴ Prob. of coin showing a tail = 1 – p

Now p_n = 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

P_n = 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) P_{n -1}+ p (1 – p) p_{n – 2} Hence Proved.

Thanks
Aditi Chauhan
askIITians Faculty

poiuhgfds · Answer

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 ​

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

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

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

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.

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