A die is tossed twice. The probability of having a number greater than 4 on each toss is

$\frac{1}{9}$

We have,

Total number of elementary events = 6 x 6 = 36

A number greater than 4 can be obtained in each toss in one of the following ways

(5,5), (5, 6), (6, 5), (6, 6)

∴ Favourable number of elementary events =4

Hence, required probability =$\frac{4}{36}=\frac{1}{9}$