A pair of unbiased dice is rolled together till a sum of either 5 or 7 is obtained.The probability that 5 comes before 7 is

With a pair of dice $P\left(5\right)=\frac{4}{36},P\left(7\right)=\frac{6}{36}$

The probability of neither 5 nor 7 is  $\frac{26}{36}$.

If $S$ is event of obtaining 5,$F$ is the event neither 5 nor 7 we get the sequence

$S,FS,FFS,FFFS$.

The desired probability $=\frac{1}{9}+\frac{13}{18}\cdot \frac{1}{9}+{\left(\frac{13}{18}\right)}^2\cdot \frac{1}{9}+\dots$

$=\frac{\frac{1}{9}}{\left(1-\frac{13}{18}\right)}\Rightarrow \frac{2}{5}$