# A fair die is tossed once. The probability that either an even number or 3 will appear

$\begin {array} {1 1} (1)\;\large\frac{4}{5} & \quad (2)\;\large\frac{3}{4} \\ (3)\;\large\frac{2}{3} & \quad (4)\;None\: of \: these \end {array}$

The favourable number of cases is 4, because getting 2,4,6 and 3
are favourable events.
Hence the required probability is $\large\frac{4}{6}$
$= \large\frac{2}{3}$
Ans : (C)
edited Aug 11, 2014