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

The correct option is B

$2/5$

Explanation for the correct option:

Step1. Assume the events :

A pair of dice is thrown so the total cases $=36$

Let $A$be the event that the sum $5$occurs, $B$be the event that the sum $7$occurs, and $C$be the event that neither the sum $5$nor sum $7$occurs.

$$\begin{gathered} A=\left\{\right(1,4),(2,3),(3,2),(4,1\left)\right\} \\ B=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\} \end{gathered}$$

$\therefore P\left(A\right)=\frac{4}{36},P\left(B\right)=\frac{6}{36}andP\left(C\right)=1-\left[\frac{4}{36}+\frac{6}{36}\right]=\frac{26}{36}$

Step2. Calculate the probability that $5$comes before $7$ :

The probability that $A$occurs before $B$

$\begin{array}{rcl}& =& P\left(A\right)+P\left(C\right).P\left(A\right)+P\left(C\right)P\left(C\right)P\left(A\right)+\dots ......\\ & =& P\left(A\right)+P\left(C\right)P\left(A\right)+P{\left(C\right)}^{2}P\left(A\right)+\dots ........\\ & =& \frac{P\left(A\right)}{[1-P(C\left)\right]}\\ & =& \frac{{\displaystyle \frac{4}{36}}}{\left[1-{\displaystyle \frac{26}{36}}\right]}\\ & =& \frac{4}{10}\\ & =& \frac{2}{5}\end{array}$

Hence the correct option is B.