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

	COLUMN - I		COLUMN - II
A)	p1	p)	1 – 2p2 + p3
B)	p2	q)	1 – p2
C)	p3	r)	1
D)	pn	s)	(1–p) pn-1+p(1–p) pn–2

	(A)	(B)	(C)	(D)
1)	p	q	r	s
2)	r	q	p	s
3)	s	p	q	r
4)	s	r	q	p

p = P(H) and P(T) = 1–P(H) = 1–p
Given Pn= Prob (getting no two consecutive heads
when the coin is tossed n times)
A) p1=P (sure event)=1 [$\because$coin is tossed only one
time it is sure that no two consecutive heads occur
B) ${\mathrm{p}}_2=\mathrm{P}\left(\mathrm{HT}\right)+\mathrm{P}\left(\mathrm{TH}\right)+\mathrm{P}\left(\mathrm{TT}\right)$
$\begin{array}{l}=1-\mathrm{P}\left(\mathrm{HH}\right)=1-\mathrm{P}\left(\mathrm{H}\right)\cdot \mathrm{P}\left(\mathrm{H}\right)=1-{\mathrm{p}}^2\\ \text{ C) }{\mathrm{p}}_3=1-\left[\mathrm{P}\right(\mathrm{HHT})+\mathrm{P}(\mathrm{THH})+\mathrm{P}(\mathrm{HHH}\left)\right]\\ =1-[\mathrm{p}\cdot \mathrm{p}\cdot (1-\mathrm{p})+(1-\mathrm{p})\cdot \mathrm{p}\cdot \mathrm{p}+\mathrm{p}\cdot \mathrm{p}\cdot \mathrm{p}]\\ =1-2{\mathrm{p}}^2+{\mathrm{p}}^3\end{array}$
D) pn=[P( Last out come is tail) x P( no two
consecutive heads occur in first (n-1) trials)]+[P(Last
out- come is head) x P((n–1)th out-come is tail).
P (no two consecutive heads occur in first (n–2)
trials)] (1 p).p_n-1 p(1 p).p_n-2