Question

$A$ and $B$ play a match in which the chances of their winning a game is $\frac{1}{4}$ for both of them and $\frac{1}{2}$ is the probability that match is being drawn. Match is finished as soon as either player wins two games. Find the probability that the match will be finished in $4$ or less games.

• A. $\frac{9}{16}$
• B. $\frac{7}{16}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{1}{2}$

We have to find the probability that match will be finished in $4$ or less games.
We will find the ways in which match is not finished in $4$ games.
There are following $4$ mutually exclusive ways:
(1) All the games are drawn.
(2) $A$ and $B$ both win one game each and remaining two are drawn.
(3) $A$ wins one game and remaining three are drawn.
(4) $B$ wins one game and remaining three are drawn.
So the probability that match is not finished in 4 games is equal to the sum of probabilities of $4$ ways.
$={\left(\frac{1}{2}\right)}^4{+}^4{C}_2{\left(\frac{1}{4}\right)}^2{\left(\frac{1}{2}\right)}^2{+}^4{C}_1\left(\frac{1}{4}\right){\left(\frac{1}{2}\right)}^3{+}^4{C}_1\left(\frac{1}{4}\right){\left(\frac{1}{2}\right)}^3$
$=\frac{1}{16}+\frac{3}{16}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$
Hence the required probability $=1-\frac{1}{2}=\frac{1}{2}$