Question

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, then

Answer 1

For $n\ge 3$, 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,
$p_n=p_{n-1}\times P$ (nth toss results in a tail)$+p_{n-2}\times P$ (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,
$p_2=1-P\left(HH\right)=1-pp=1-p^2$.
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) $p_n=(1-p)p_{n-1}+p(1-p)p_{n-2}$