A die has 6 faces marked by the given numbers as shown below :

$\boxed{\phantom{1}1\phantom{1}}\space\boxed{\phantom{1}2\phantom{1}}\space\boxed{\phantom{1}3\phantom{1}}\space\boxed{-1}\space\boxed{-2}\space\boxed{-3}$

The die is thrown once. What is the probability of getting

(i) a positive integer

(ii) an integer greater than -3

(iii) the smallest integer ?

On a single throw of die,

Sample space = {1, 2, 3, -1, -2, -3}.

(i) Let E₁ be the event of getting a positive integer, then

E₁ = {1, 2, 3}.

∴ The number of favourable outcomes to the event E₁ = 3.

$\therefore P(E$1) = \dfrac{\text{No. of favourable outcomes to } E1}{\text{Total no. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}.

Hence, the probability of getting a positive integer is $\dfrac{1}{2}$.

(ii) Let E₂ be the event of getting a integer greater than -3, then

E₂ = {-2, -1, 1, 2, 3}.

∴ The number of favourable outcomes to the event E₂ = 5.

$\therefore P(E$2) = \dfrac{\text{No. of favourable outcomes to } E2}{\text{Total no. of possible outcomes}} = \dfrac{5}{6}.

Hence, the probability of getting a integer greater than -3 is $\dfrac{5}{6}$.

(iii) Let E₃ be the event of getting smallest integer, then

E₃ = {-3}.

∴ The number of favourable outcomes to the event E₃ = 1.

$\therefore P(E$3) = \dfrac{\text{No. of favourable outcomes to } E3}{\text{Total no. of possible outcomes}} = \dfrac{1}{6}.

Hence, the probability of getting smallest integer is $\dfrac{1}{6}$.