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

Options

A. `(1)/(3)`
B. `(1)/(9)`
C. `(2)/(3)`
D. `(1)/(12)`

Answer

B

Explanation

We have,
Total number of elementary events `=6xx6=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)
`therefore` Favourable number of elementary events =4
Hence, required probability `=(4)/(36)=(1)/(9)`