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Q.

A and B throw a pair of dice. A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A throws 6. If A begins the game his chance of winning is

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a

$\frac{5}{61}$

b

$\frac{30}{61}$

c

$\frac{35}{61}$

d

$\frac{60}{61}$

answer is B.

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Detailed Solution

$\begin{array}{l} \bar{a} = 1 - \frac{5}{36} = \frac{31}{36} \bar{b} = 1 - \frac{1}{6} = \frac{5}{6} \\ a = \frac{5}{36} b = \frac{1}{6} \end{array} \begin{array}{l} P (A w i n) = a + \bar{a} \bar{b} \bar{a} \bar{b} a + - - - - - - \infty \\ = \frac{a}{1 - \bar{a} \bar{b}} = \frac{30}{61} \end{array}$

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