Mathematics - Functions Question with Solution | TestHub

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. $\begin{array}{rl}& \therefore \mathrm{m}=6\\ & \mathrm{n}=\text{ Total number of cases }=6\times 6\end{array}$ ( $\therefore$ with each die there are six possibilities) $\therefore$ Probability $\mathrm{p}=\frac{\mathrm{m}}{\mathrm{n}}=\frac{6}{6\times 6}=\frac{1}{6}$