If two loaded dice each have the property that 2 or 4 is three times as likely to appear as 1, 3, 5 or 6 on each roll. When two such dice are rolled, the probability of obtaining a total of 7 is p then the value of $\left[1/p\right]$is (where [.] represents the greatest integer function)

$\text{ Let the probability of the faces }1,3,5\text{ or }6\text{ be }\mathrm{p}\text{ for each face. }$

$\text{ Hence probability of each of the faces }2\text{ or }4\text{ is }3p$

$\text{ Now according to the question }4p+6p=1$

$\begin{array}{l}\Rightarrow p=\frac{1}{10}\\ \therefore P\left(1\right)=P\left(3\right)=P\left(5\right)=P\left(6\right)=\frac{1}{10};\\ \text{ and }P\left(2\right)=P\left(4\right)=\frac{3}{10}\end{array}$

Required probability 

$p=P(\text{ total of }7\text{ with a draw of dice) }$

$\begin{array}{l}p=P\left(\text{ total of }7\text{ with a draw of dice }\right)\\ =P(16,61,25,52,43,34)\\ =2\left(\frac{1}{10}\cdot \frac{1}{10}\right)+2\left(\frac{3}{10}\cdot \frac{1}{10}\right)+2\left(\frac{3}{10}\cdot \frac{1}{10}\right)\\ =\frac{2+6+6}{100}=\frac{14}{100}=\frac{7}{50}\end{array}$