Question

A player tosses a coin and scores one point for every head and two point for every tail that truns up. He plays on until his scores reaches or psses $n$. $P_n$ denotes the probability of getting a scores of exactly $n$
$$\begin{array}{cc}\text{List I}& \text{List II}\\ \begin{array}{cc}\text{(a)}& \text{the value of}{P}_n\text{is}\end{array}& \text{(p)}1\\ \begin{array}{c}\text{(b) the value of}{P}_n+\frac{1}{2}{P}_{n-1}\end{array}& \text{(q)}\frac{5}{4}\\ \begin{array}{c}\text{(c)}2P_{101}+P_{100}\end{array}& \text{(r)}2\\ \begin{array}{c}\text{(d)}{P}_1+P_2\end{array}& \text{(s)}\frac{1}{2}[P_{n-1}+P_{n-2}]\end{array}$$
$\text{Which of the following is the only}\text{correct option?}$

• A. $\left(a\right)\to \left(s\right)$
• B. $\left(b\right)\to \left(r\right)$
• C. $\left(c\right)\to \left(q\right)$
• D. $\left(d\right)\to \left(r\right)$

Answer 1

The correct option is A $\left(a\right)\to \left(s\right)$

The scores of $n$ can be reached in the following two mutually exlcusive events
$a:$by throwing a head when the score is $(n-1)$
$b:$by throwing a tail when the score is $(n-2)$
$P_n=\frac{1}{2}[P_{n-1}+P_{n-2}]$
$P_n+\frac{1}{2}{P}_{n-1}=P_{n-1}+\frac{1}{2}{P}_{n-2}\Rightarrow P_n+\frac{1}{2}{P}_{n-1}=P_2+\frac{1}{2}{P}_1$
Now $P_1=\frac{1}{2}$
For $P_2=\frac{1}{2}+\frac{1}{2}\times \frac{1}{2}=\frac{3}{4}\to P_n+\frac{1}{2}{P}_{n-1}=\frac{3}{4}+\frac{1}{2}\times \frac{1}{2}=1$
$P_n-\frac{2}{3}=1-\frac{1}{2}{P}_{n-1}-\frac{2}{3}=-\frac{1}{2}(P_{n-1}-\frac{2}{3})\to P_{n-1}-\frac{2}{3}=-\frac{1}{2}(P_{n-2}-\frac{2}{3})\therefore P_n-\frac{2}{3}=\frac{1}{(-2{)}^{n-1}}(P_1-\frac{2}{3})P_n=\frac{2}{3}+\frac{(-1{)}^n}{2^n\times 3}{P}_{100}=\frac{2}{3}+\frac{1}{3\times 2^{100}}{P}_{101}=\frac{2}{3}-\frac{1}{3\times 2^{101}}{P}_{100}+2P_{100}=2$