Question

A die is tossed twice. Find the probability of getting a number greater than 3 on each toss.

Answer 1

$$\begin{gathered} S=\{\left(11\right),\left(12\right),\left(13\right),\left(14\right),\left(15\right),\left(16\right), \\ \left(21\right),\left(22\right),\left(23\right),\left(24\right),\left(25\right),\left(26\right), \\ \left(31\right),\left(32\right),\left(33\right),\left(34\right),\left(35\right),\left(36\right), \\ \left(41\right),\left(42\right),\left(43\right),\left(44\right),\left(45\right),\left(46\right), \\ \left(51\right),\left(52\right),\left(53\right),\left(54\right),\left(55\right),\left(56\right), \\ \left(61\right),\left(62\right),\left(63\right),\left(64\right),\left(65\right),\left(66\right)\} \\ n\left(\mathrm{S}\right)=36 \\ E=\text{Getting a number greater than 3 on each toss} \\ =\left\{\left(44\right),\left(45\right),\left(46\right),\left(54\right),\left(55\right),\left(56\right),\left(64\right),\left(65\right),\left(66\right)\right\} \\ n\left(\mathrm{E}\right)=9 \\ P\left(E\right)=\frac{9}{36} \\ =\frac{1}{4} \end{gathered}$$