Probability Ques 77
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:
Formula:
- 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.
BETA
