Question

What is the chance of throwing a number greater than $4$ with an ordinary die whose faces are numbered from $1$ to $6$$?$

Answer 1

There are $6$ possible ways in which the die can fall, and of these two are favourable to the event required i.e. getting a number greater than $4$ which is $5$ and $6$.
Therefore the required chance $=\frac{2}{6}=\frac{1}{3}$.