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 1/pis (where [.] represents the greatest integer function)

Let the probability of the faces 1,3,5 or 6 be p for each face.  Hence probability of each of the faces 2 or 4 is 3p Now according to the question 4p+6p=1⇒p=110∴ P(1)=P(3)=P(5)=P(6)=110; and P(2)=P(4)=310Required probability p=P( total of 7 with a draw of dice) p=P( total of 7 with a draw of dice )=P(16,61,25,52,43,34)=2110⋅110+2310⋅110+2310⋅110=2+6+6100=14100=750