SATHEE: Probability 3 Question 37

Probability 3 Question 37

37. A coin has probability $p$ of showing head when tossed. It is tossed $n$ times. Let $p_{n}$ denotes the probability that no two (or more) consecutive heads occur. Prove that $p_{1} = 1$ , $p_{2} = 1 - p^{2}$ and $p_{n} = (1 - p) \cdot p_{n - 1} + p (1 - p) p_{n - 2}, \forall n \geq 3$ .

$(2000, 5 M)$

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Solution:

Since, $p_{n}$ denotes the probability that no two (or more) consecutive heads occur.

$\Rightarrow p_{n}$ denotes the probability that 1 or no head occur.

For $n = 1, p_{1} = 1$ because in both cases we get less than two heads $(H, T)$ .

For $n = 2, p_{2} = 1 - p$ (two heads simultaneously occur).

$= 1 - p (H H) = 1 - p p = 1 - p^{2}$

For $n \geq 3, p_{n} = p_{n - 1} (1 - p) + p_{n - 2} (1 - p) p$

$\Rightarrow p_{n} = (1 - p) p_{n - 1} + p (1 - p) p_{n - 2}$

Hence proved.

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Probability 3 Question 37

37. A coin has probability p of showing head when tossed. It is tossed n times. Let pn denotes 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,∀n≥3.

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37. A coin has probability $p$ of showing head when tossed. It is tossed $n$ times. Let $p_{n}$ denotes the probability that no two (or more) consecutive heads occur. Prove that $p_{1} = 1$ , $p_{2} = 1 - p^{2}$ and $p_{n} = (1 - p) \cdot p_{n - 1} + p (1 - p) p_{n - 2}, \forall n \geq 3$ .