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 geq 3`.

Explanation

Given that the probability of showing head by a coin when tossed is p.
So, the probability of coin not showing head is (1-p). Now, `p_(n)` denotes probability that no two or more consecutive heads occur in n throws.
Clearly, `p_(1)=1` as when coin is tossed once there will be no two consectiven heads.
Also, `p_(2)=P(HT)+P(TH)+P(T T)`
`=p(1-p)+p(1-p)+(1-p)^(2)=1-p^(2)`
Let event A is "last toss is tail" and evetn B is "last toss is head and second last toss is tail."
`therefore` Using total probability theorem,
`p_(n)=p_(n-1)xxP(A)+p_(n-2)xxP(B)`
`thereforep_(n)=(1-p)p_(n-1)+p(1-p)p_(n-2)"for all n"ge3.`