NIMCET 2008 Question Paper with solutions

Number of cases for sum '5' $=4$
Number of cases for sum '7' = 6
Probability of geting sum 5 in one roll $=\frac{4}{36}=\frac{1}{9}$
Probability of getting either 5 or 7 in a roll
$=\frac{4+6}{36}=\frac{5}{18}$
$\Rightarrow$ Probability of getting 5 before 7
$=\frac{1}{9}+\frac{13}{18}\cdot \frac{1}{9}+{\left(\frac{13}{18}\right)}^2\cdot \frac{1}{9}+...$
$=\frac{\frac{1}{9}}{1-\frac{13}{18}}=\frac{1}{9}\times \frac{18}{5}=\frac{2}{5}$