Question : A and B play a game of tennis. The situation of the game is as follows; if one scores two consecutive points after a deuce, he wins, if loss of a points is followed by win of a point, it is deuce. The chance of a server to win a point is 2/3. The game is at deuce and A is serving. Probability that A will win the match is (Serves are changed after each game)
(A) 3/5
(B) 2/5
(C) 1/2
(D) 4/5
Answer : option (C)
Solution :
Let us assume that A wins after n deuces, n = 0, 1, 2, 3….. The probability of deuce is (2/3) × (2/3) + (1/3) × (1/3) = 5/9. [A wins his serve, then B wins his serve or A loses his serve].
So, the probability that A wins the game after ‘n’ deuces is \left( \dfrac{5}{9}\right) ^{n}\times \left( \dfrac{2}{3}\right) \times \left( \dfrac{1}{3}\right) [After nth deuce A serves and wins, then B serves and loses]. Therefore, the required probability of ‘ A’ winning the game is
\begin{aligned}\sum ^{\infty }_{n=0}\left( \dfrac{5}{9}\right) ^{n}\times \dfrac{2}{3}\times \dfrac{1}{3}\ \Rightarrow \dfrac{1}{1-\dfrac{5}{9}}\times \dfrac{2}{9}\Rightarrow \dfrac{1}{2}\end{aligned}