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  • A,B and C throw a pair of dice in that order alternately till one of them gets a total of 9 and wins the game. Find their respective probabilities of winning, if A starts first.     </d

A,B and C throw a pair of dice in that order alternately till one of them gets a total of 9 and wins the game. Find their respective probabilities of winning, if A starts first. 

 

 

 

 
 
 
 
 

Answers (1)

Let E : getting a total of 9

\therefore E=\left \{ (3,6),(4,5),(5,4),(6,3) \right \}

So, clearly the probability if winning :

P(E)=\frac{4}{36}=\frac{1}{9}

\therefore P(\bar{E})=\frac{8}{9}

As A starts the game so he may have win in 1st, 4th, 7th

\therefore P(A \; wins)=P(E)+P(\bar{E})P(\bar{E})P(\bar{E})P(E)+P(\bar{E})P(\bar{E})P(\bar{E})P(\bar{E})P(\bar{E})P(\bar{E})P(E)+...

\Rightarrow P(A \; wins)=\frac{1}{9}+\left ( \frac{8}{9} \right )^3\times \frac{1}{9}+\left ( \frac{8}{9} \right )^6\times \frac{1}{9}+...

\therefore P(A \; wins)=\frac{\frac{1}{9}}{1-\frac{512}{729}}=\frac{81}{217}

Now B may win in 2nd, 5th, 8th... trials 

\therefore P(B \; wins)=P(\bar{E})P(E)+P(\bar{E})P(\bar{E})P(\bar{E})P(\bar{E})P(E)+...

\Rightarrow P(B \; wins)=\frac{8}{9}\times \frac{1}{9}+\left ( \frac{8}{9} \right )^4\times \frac{1}{9}+...

\therefore P(B \; wins)= \frac{\frac{8}{81}}{1-\frac{512}{729}}=\frac{72}{217}

And finally P(C \; wins)= 1-\left [ P(A\; wins)+P\left ( B\; wins \right ) \right ]

                         = 1-\left [ \frac{81}{217}+\frac{72}{217}\right ]= \frac{64}{217}

 

Posted by

Ravindra Pindel

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