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}. $