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 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.
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.
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