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In a hockey match,both teams A and B scored same number of goals up to the end of the game,so to decide the winner ,the referee asked both the captains to throw a die alternately and decided that the team,whose captain gets a six first,will be declared the winner.If the captain of team A was asked to start,find their respective probabilities of winning the match and state whether the decision of the referee was fair or not.

This question appeared in the 65-1,65-2 and 65-3 versions of the paper in 2013.
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Let the probability that A gets a six be $\frac{1}{6}$
$\Rightarrow$ $p=\frac{1}{6}$ and hence $q=1-\frac{1}{6} =\frac{5}{6}$
Probability of A's winning $=P(A)+P(\bar {A} \bar {B} A)+P(\bar {A} \bar {B} \bar {A} \bar {B}A)+.....$
This is in geometric progression, where summation is $S_n =\frac{P}{1-q^2}$
Substituing for P and q, $\large\frac{1/6}{1-(5/6)^2}=\frac{1/6}{\frac{36-25}{36}}$
Therefore The required probability of winning is P(A's winning)=$\large\frac{6}{11}$
and P(B's winning)=$1-\large\frac{6}{11}=\frac{5}{11}$
answered Mar 21, 2013 by meena.p
edited Mar 26, 2013 by balaji.thirumalai

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