Question

A pair of dice is rolled together till a sum of either $5$ or $7$ is obtained. The probability that $5$ comes before $7$ is $k/10$.Find $k$

Answer 1

$5$ can be thrown in $4$ ways and $7$ can be thrown in $6$ ways, hence number of ways of throwing neither $5$ nor $7$ is
$36-(4+6)=26$
Therefore probability of throwing a five in a single throw with a pair of dice is $\frac{4}{36}=\frac{1}{9}$ and probability of throwing neither $5$ nor $7$ is $\frac{26}{36}=\frac{13}{18}$
Hence, required probability $=\left(\frac{1}{9}\right)+\left(\frac{13}{18}\right)\left(\frac{1}{9}\right)+{\left(\frac{13}{18}\right)}^2\left(\frac{1}{9}\right)+...=\frac{\frac{1}{9}}{1-\frac{13}{18}}=\frac{2}{5}$