Question

In a game, winning is defined by the outcome $6$ on a die. Three players $A,B,C$ play this game. The die is first thrown by $C$, followed by $A$ and $B$. If the order of throwing the die is same for all rounds, then the probability that $A$ wins the game, is

• A. $\frac{60}{91}$
• B. $\frac{25}{91}$
• C. $\frac{66}{91}$
• D. $\frac{30}{91}$

Answer 1

The correct option is D $\frac{30}{91}$

The order is $C,A,B$.
So the probability that $A$ wins, is
$\frac{5}{6}\cdot \frac{1}{6}+{\left(\frac{5}{6}\right)}^2\frac{5}{6}\cdot \frac{5}{6}\cdot \frac{1}{6}+\cdots =\frac{\frac{5}{36}}{1-\frac{5^3}{6^3}}=\frac{30}{91}$