A black die and a white die are rolled. Find the probability that...

The number of favourable cases are shown below: $$\begin{array}{cl}\text{ Number on white die }& \text{ Number on black die }\\ 1 & 3 \\ 1 & 4 \\ 1 & 5 \\ 1 & 6 \\ 2 & 5 \\ 2 & 6\end{array}$$) There are 6 favourable cases in which the number on black die is more than twice the number on the white die. $$\begin{aligned}& \therefore \mathrm{m}=6 \\ & \mathrm{n}=\text{ Total number of cases }=6 \times6 \end{aligned}$$) ( (\therefore) with each die there are six possibilities) (\therefore) Probability \$(\mathrm{p}=\frac{\mathrm{m}}{\mathrm{n}}=\frac{6}{6 \times 6}=\frac{1}{6}$)