A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is [2025]

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Solution

Q.
A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is [2025]

1 $\frac{8}{17}$

2 $\frac{9}{19}$

3 $\frac{9}{17}$

4 $\frac{8}{19}$

Ans.
(2)

Let $E_{1}$ be the event that A get sum of 5 and $E_{2}$ be the event that B get sum of 8.

$P (E_{1}) = \frac{4}{36} = \frac{1}{9} and P (E_{2}) = \frac{5}{36}$

$∴$ Required Probability

$= P (E_{1}) + P ({\bar{E}}_{1}) P ({\bar{E}}_{2}) P (E_{1}) + P ({\bar{E}}_{1}) P ({\bar{E}}_{2}) P ({\bar{E}}_{1}) P ({\bar{E}}_{2}) P (E_{1}) + . . . \infty$

$= \frac{1}{9} + \frac{8}{9} \times \frac{31}{36} \times \frac{1}{9} + \frac{8}{9} \times \frac{31}{36} \times \frac{8}{9} \times \frac{31}{36} \times \frac{1}{9} + . . . \infty$

$= \frac{\frac{1}{9}}{1 - \frac{62}{81}} = \frac{9}{19}$ .

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