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, $D D, L D, D L$ are possibilities for remaining 2 matches. Hence, the required probability
$={ }^{3} C _{1}\left(\frac{1}{7}\right)\left(\frac{2}{7} \times \frac{2}{7}+\frac{2}{7} \times \frac{4}{7}+\frac{4}{7} \times \frac{2}{7}\right) \times \frac{1}{7}$
$=\frac{60}{7^{4}}=\frac{60}{2401}=p$
$\Rightarrow 343 p=\frac{60}{7}$