Question

A chess match between $2$ players $A$ and $B$ is won by whoever first wins a total of $2$ games. Probability of $A$ winning, drawing and losing any particular game is $\frac{1}{7},\frac{2}{7}$ and $\frac{4}{7}$ respectively (the games are independent). If the probability that $A$ wins the match in the fourth game is $p$ , then $343p$ is equal to

• A. $60$
• B. $30$
• C. $\frac{60}{7}$
• D. $\frac{30}{7}$

Answer 1

Answer: (C)

A win match in $4^{\text{th }}$ game $\Rightarrow A$ win one game in first 3 games, $DD,LD,DL$ are possibilities for remaining 2 matches. Hence, the required probability
$={}^3{C}_1\left(\frac{1}{7}\right)\left(\frac{2}{7}\times \frac{2}{7}+\frac{2}{7}\times \frac{4}{7}+\frac{4}{7}\times \frac{2}{7}\right)\times \frac{1}{7}$
$=\frac{60}{7^4}=\frac{60}{2401}=p$
$\Rightarrow 343p=\frac{60}{7}$