Question

The probability that $X$ wins the match after $(n+1)$ games, $(n\ge 1)$ is

• A. $n{a}^2{b}^{n-1}$
• B. $a^2(n{b}^{n-1}+n(n-1\left)b^{n-2}c\right)$
• C. $n{a}^{2}b{c}^{n-1}$
• D. none of these

Answer 1

The correct option is B $a^2(n{b}^{n-1}+n(n-1\left)b^{n-2}c\right)$

$X$ can win after the $(n+1{)}^{th}$ game in the following two mutually exclusive ways
(i) $X$ wins exactly one of the first $n$ games draws $(n-1)$ games and wins the $(n+1{)}^{th}$ the games
(ii) $X$ loses exactly one of the first $n$ games, wins exactly one of the first $n$ games and draws $(n-2)$ games and wins the $(n+1{)}^{th}$ game.
For (i) the probability is ${(}^n{P}_{1}a{b}^{n-1})a$ and for (ii) the probability is ${(}^n{P}_2\left)\right(ac)b^{n-2}a$
$\therefore$ The probability $X$ wins after the $(n+1{)}^{th}$ game
$P\left(X\right)=n{a}^2{b}^{n-1}+n(n-1)a^2{b}^{n-2}c$
$=a^2(n{b}^{n-1}+n(n-1\left)b^{n-2}c\right)$.