Question

A pair of fair dice is rolled together till a sum of either $5$ or $7$ is obtained. Then the probability that $5$ before $7$ is

Answer 1

Probability of getting a sum of $5$ is $\frac{4}{36}=\frac{1}{9}=P\left(A\right)$ as favorable cases are $\left\{\right(1,4),(4,1),(2,3),(3,2\left)\right\}$. Similarly, favorable cases of getting a sum of $7$ are $\left\{\right(1,6),(6,1),(2,5),(5,2),(3,4).(4,3\left)\right\}$. The total number of such cases is $6$. Therefore, the probability of getting a sum of $5$ or $7$ is $\frac{1}{6}+\frac{1}{9}=\frac{5}{18}$(as events are mutually exclusive)
Hence, the probability of geting neither a sum of $5$ nor a sum of $7$ is $1-\frac{5}{18}=\frac{13}{18}$.
Now, we get a sum of $5$ before a sum of $7$ if either we get a sum of $5$, in first chance or we get neither a sum of $5$ nor a sum of $7$ in first chance and a sum of $5$ in second chance and so on.
Therefore the recquired probability is
$\frac{1}{9}+\frac{13}{18}\times \frac{1}{9}+\frac{13}{18}\times \frac{13}{18}\times \frac{1}{9}+\cdots +\infty =\frac{1/9}{1-13/18}=\frac{2}{5}$