In a single throw of a die, find the probability of getting:
(i) 8
(ii) a number greater than 8
(iii) a number less than 8

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Solution

On a die the numbers are 1, 2, 3, 4, 5, 6 i.e. six.
$∴$ Number of possible outcome = 6
(i) Number of favourable outcome = 0 ( $∵$ 8 is not possible )
$∴$ P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e}{N u m b e r o f a l l p o s s i b l e o u t c o m e}$
$= \frac{0}{6} = 0$
(ii) Number greater than 8 will be 0
$∴$ P(E)= $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e}{N u m b e r o f a l l p o s s i b l e o u t c o m e}$
$= \frac{0}{6} = 0$
(iii) Number less than 8 will be 1, 2, 3, 4, 5, 6
$∴$ P(E) = $\frac{N u m b e r o f f a v o u r a b l e o u t c o m e}{N u m b e r o f a l l p o s s i b l e o u t c o m e}$
$= \frac{6}{6} = 1$

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