Question

A chess game between $X$ and $Y$ is won by whoever first wins a total of $2$ games.

$X’s$ chances of winning, drawing, or losing a particular game are $\frac{1}{6},\frac{1}{3},\frac{1}{2}$.

The games are independent.

The probability that $Y$ winning that match in the $4th$ game is

• A. $\frac{1}{6}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is A

$\frac{1}{6}$

The explanation for the correct option:

Step1. The given probability :

Given $X’s$chances of winning $=\frac{1}{6}$

$\Rightarrow$ $Y’s$chances of losing $=\frac{1}{6}$

and $X’s$chances of drawing $=\frac{1}{3}$

$\Rightarrow$ $Y’s$chances of drawing $=\frac{1}{3}$

and $X’s$chances of losing $=\frac{1}{2}$

$\Rightarrow$ $Y’s$chances of winning $=\frac{1}{2}$

Step2. Calculate the probability of $Y$wins :

The probability that the $Y$wins this match in $4th$ game is

$=$$P$[ ( $X$ win)(draw)( $Y$ win)( $Y$ win)]$\times 3!+{}^{3}C_1$ $P$[ (draw)(draw)( $Y$win)( $Y$ win)]

$$\begin{gathered} =\left[\left(\frac{1}{6}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{2}\right)\times \left(\frac{1}{2}\right)\times 6\right]+{}^{3}C_1\left[\left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{2}\right)\times \left(\frac{1}{2}\right)\right] \\ =\frac{6}{72}+\frac{3}{36} \\ =\frac{12}{72} \\ =\frac{\mathbf{1}}{\mathbf{6}} \end{gathered}$$

Hence, the correct option is (A).