Question

In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will be an old number.
(iii) will not be an even number.

Answer 1

A die has six numbers: 1, 2, 3, 4, 5, 6
$\therefore$ Number of possible outcome= 6
(i) Number of favourable outcome= an even number i.e. 2, 4, 6 which are 3 in numbers
$\therefore$ P(E) = $\frac{Numberoffavourableoutcome}{Numberofallpossibleoutcome}$
=$\frac{3}{6}=\frac{1}{2}$
(ii) and (iii) Number of favourable outcome = not an even number i.e. odd numbers : 1, 3, 5 which are 3 in numbers
$\therefore$ = $\frac{Numberoffavourableoutcome}{Numberofallpossibleoutcome}$
$=\frac{3}{6}=\frac{1}{2}$