# In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores

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In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six The game stops as soon as either of the players wins. The probability of A winning the game is :

(1) 31/61

(2) 5/6

(3) 5/31

(4) 30/61

by (47.6k points)
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Answer is (4) $\frac{31}{61}$

P(A) = $\frac{5}{36}$, P(B) = $\frac{1}{6}$

P(A wins) = W + FFW + FFFFW + .........

= $\frac{5}{36} + \frac{31}{36} \times \frac{5}{6} \times \frac{5}{36}$+ $\frac{31}{36} \times \frac{5}{6} \times \frac{31}{36} \times \frac{5}{6} \times \frac{5}{36} + ........$

= $\frac{\frac{5}{36}}{1-\frac{31}{36} \times \frac{5}{6}}$

= $\frac{\frac{5}{36}}{\frac{36 \times 6 - 31 \times 5}{36 \times 6}}$

= $\frac{30}{216-155}$

= $\frac{31}{61}$

+1 vote
by (20 points)
P(6)=5/36

P(7)=6/36=1/6

P(A wins)=1/6 + 31/36 * 5/6 * 1/6 ..........................

=1/6[1/(1-155/216)]

= 30/61