GATE 2015 EE - SET 1 -Question 29

Solution :

Probability of getting a 6 = \frac{1}{6}

Probability of A Winning is  = \frac{1}{6}

Probability of A losing is = 1 - \frac{1}{6}\frac{5}{6}

Probability of B winning is = \frac{1}{6}

Probability of  B losing is = \frac{1}{6}

If A starts the game , the probability of A winning ,

P(A) + P(\bar{A})P(\bar{B})P(A)+P(\bar{A})P(\bar{B})P(\bar{A})P(\bar{B})P(A)

\frac{1}{6} + \frac{5}{6}* \frac{5}{6} *\frac{1}{6} +\frac{5}{6}* \frac{5}{6}* \frac{5}{6}* \frac{5}{6}* \frac{1}{6}* + ...................

\frac{1}{6}\left [ 1+ \frac{5}{6}* \frac{5}{6}+\frac{5}{6}* \frac{5}{6}*\frac{5}{6}* \frac{5}{6}+....\right ]

\frac{1}{6}\left [1 + (\frac{5}{6}) ^{2}+(\frac{5}{6}) ^{4}+ .... \right ]

\left [ 1 + a^{2} +a^{4}+....\right ]

\Rightarrow \left [ \frac{1}{1-a^{2}} \right ]

a = \frac{1}{6} , r = (\frac{5}{6})^{2} = \frac{25 }{36}

a = \frac{1}{6} , r = a = \frac{25}{36}

S_{\infty } = \frac{\frac{1}{6}}{1-\left ( \frac{25}{36} \right )}

\frac{\frac{1}{6}}{(\frac{36-25}{36})} =  \frac{6 }{11}