Question

A die is thrown, find the probability of getting:
(i) a prime number
(ii) a number greater than
(iii) a number not greater than 4.

Answer 1

A die has six numbers: 1, 2, 3, 4, 5, 6
$\therefore$ Number of possible outcomes = 6
(i) a prime number
Number of favourable outcomes= a prime number= 1, 3, 5 which are 3 in numbers
P(E)= $\frac{Numberoffavourableoutcome}{Numberofallpossibleoutcome}$
$=\frac{3}{6}=\frac{1}{2}$
(ii) Number of favourable outcome = greater than four i.e. two numbers 5 and 6 $\therefore$ P(E) = $\frac{Numberoffavourableoutcome}{Numberofallpossibleoutcome}$
$\frac{2}{6}=\frac{1}{3}$
(iii) Number of favourable outcome = not greater than 4 or numbers will be 1, 2, 3, 4 which are 4 in numbers
$\therefore$ P(E) = $\frac{Numberoffavourableoutcome}{Numberofallpossibleoutcome}$
$=\frac{4}{6}=\frac{2}{3}$