Question

Two persons A and B throw a die alternately till one of them gets a $3$ and wins the game, the respective probabilities of winning, if $A$ begins, are:

• A. $\frac{7}{11},\frac{4}{11}$
• B. $\frac{6}{11},\frac{5}{11}$
• C. $\frac{5}{6},\frac{1}{6}$
• D. $\frac{4}{7},\frac{3}{7}$
• E. $\frac{1}{2},\frac{1}{2}$

Answer 1

Answer: (B)

$\because$ $p(A)=\frac{1}{6},p(\overline{A})=\frac{5}{6}$
and $p(B)=\frac{1}{6},p(\overline{B})=\frac{5}{6}$
Hence, Probability of winning of A $[=P(E)+P(\overline{E}\cap \overline{F}\cap \overline{E})+$
$P(\overline{E}\cap \overline{F}\cap \overline{E}\cap \overline{F}\times E)+....$
$=\frac{1}{6}+{\left(\frac{5}{6}\right)}^2\left(\frac{1}{6}\right)+{\left(\frac{5}{3}\right)}^4\left(\frac{1}{6}\right)+.....$ $=\frac{\frac{1}{6}}{1-{\left(\frac{5}{6}\right)}^2}=\frac{6}{11}$
Also, probability of winning $B=1-\frac{6}{11}\text{=}\frac{5}{11}.$