The probability that X wins the match after ( n + 1 ) games, \[n \geqslant 1\]is
A. \[n{a^2}{b^{n - 1}}\]
B. \[{a^2}(n{b^{n - 1}} + n(n - 1){b^{n - 2}}c\]
C. \[n{a^2}b{c^{n - 1}}\]
D. \[none{\text{ }}of{\text{ }}these\]

Hint: It is possible that X will win a match of first n games or X will lose the match or the match will draw but here we had to solve the probability that X will win a match so let the probability of winning the match be P. And we all know that \[{(^n}{C_r}) = \dfrac{{n!}}{{r!}}(n - r)!\].

Complete step-by-step answer:
It is clear that X will win the match after the ( n + 1 ) game.
And this is possible by two mutually exclusive ways and these 2 ways also have three cases.

( 1 ) The 1st way of winning the match is \[{P_1}\].
Case (a) X wins exactly one of the first n games
Case (b) draw ( n – 1 ) games
Case (c) wins the ( n + 1 ) the games.
So, the probability of this way is \[{(^n}{P_1}a{b^{n - 1}})a\].

( 2 ) Now let the 2nd way of winning the match is \[{P_2}\].
Case (a) X loses exactly one of the first n games, wins exactly one of the first n games
Case (b) draws ( n – 2 ) games.
Case (c) wins the ( n + 1 ) game.
So, the probability of this way is \[{(^n}{P_2})(ac){b^{n - 2}}a\].
Now the probability X wins the match after ( n + 1 ) game is
P (X) = \[{P_1} + {P_2}\]
Now, put the value of \[{P_1}\] and \[{P_2}\] in the above equation.
P ( X ) = \[{(^n}{P_1}a{b^{n - 1}})a\]+ \[{(^n}{P_2})(ac){b^{n - 2}}a\] \[\left[ {{(^n}{P_2}) = n(n - 1)} \right]\]
P ( X ) = \[n{a^2}{b^{n - 1}} + \;n(n - 1){a^2}{b^{n - 2}}c\]
P ( X ) = \[{a^2}\left[ {n{b^{n - 1}} + \;n(n - 1){b^{n - 2}}c} \right]\]
So, the probability X wins the match after ( n + 1 ) games.
Hence B is the correct option.

Note: whenever we come up with this type of problem then we must focus on all the possible ways that can be made by the given statement. And then we have to find the probability for all the possible cases and in the end we have to add up all these to give us the desired result.