A coin shows heads with probability p. In n tosses, pn is the probability of no two consecutive heads. Prove p₁=1, p₂=1-p², the recurrence pn=(1-p)pn-1+p(1-p)pn-2, and (by induction) that pn=Aαⁿ+Bβⁿ.

Answer: Proved. Conditioning on the last toss gives the linear recurrence; its characteristic equation t²-(1-p)t-p(1-p)=0 has roots α,β, so pn=Aαⁿ+Bβⁿ by induction. JEE Advanced 2000 · Binomial Theorem — verified solution by the Vidaara Team.

JEE PYQ

[JEE Advanced 2000] a coin shows heads with probability p in n tosses p n is the

VAVidaara Admin Asked 2d ago 0 views 1 answer

#PYQJEE #JEE #JEE Advanced #2000 #mathematical-induction #Binomial Theorem

A coin shows heads with probability $p$. In $n$ tosses, $p_n$ is the probability of no two consecutive heads. Prove $p_1=1,\ p_2=1-p^2$, the recurrence $p_n=(1-p)p_{n-1}+p(1-p)p_{n-2}$, and (by induction) that $p_n=A\alpha^n+B\beta^n$.

1 Answer

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

Answer: Proved.

Conditioning on the last toss gives the linear recurrence; its characteristic equation $t^2-(1-p)t-p(1-p)=0$ has roots $\alpha,\beta$, so $p_n=A\alpha^n+B\beta^n$ by induction.

JEE Advanced 2000 · Binomial Theorem — verified solution by the Vidaara Team.

Discussion (0)

No comments yet — start the discussion.

← Back to all questions