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) \[\frac{1}{2},\frac{1}{2}\]
Correct Answer: B
Solution :
\[\because \] \[p(A)=\frac{1}{6},p(\overline{A})=\frac{5}{6}\] 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\[B=1-\frac{6}{11}\text{=}\frac{5}{11}.\]
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