• # question_answer Three persons $A,\,\,\,B,\,\,\,C$ throw a die in succession. The one getting 'six' wins. If $A$ starts then the probability of $B$ winning is A) $\frac{36}{91}$                                              B) $\frac{25}{91}$C) $\frac{41}{91}$                                              D) $\frac{30}{91}$

$P(\bar{E}E)+P(\bar{E}\bar{E}\bar{E}\bar{E}\bar{E})$                 $=\frac{5}{6}\times \frac{1}{6}+{{\left( \frac{5}{6} \right)}^{4}}\times \frac{1}{6}+{{\left( \frac{5}{6} \right)}^{8}}\frac{1}{6}...\infty$                 $=\frac{5}{36}\left[ 1+{{\left( \frac{5}{6} \right)}^{3}}... \right]=\frac{30}{91}$