A die is thrown once. What is the probability of getting:

(i) an odd number

(ii) a number greater than 4

(iii) a number less than 5

(iv) the number 5

In a single throw of die,

Sample space = {1, 2, 3, 4, 5, 6}.

(i) Let A be the event of getting an odd number, then

A = {1, 3, 5}.

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

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{6} = \dfrac{1}{2} .$

Hence, the probability of getting an odd number is $\dfrac{1}{2}.$

(ii) Let B be the event of getting a number greater than 4, then

B = {5, 6}.

∴ The number of favourable outcomes to the event B = 2.

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{6} = \dfrac{1}{3} .$

Hence, the probability of getting a number greater than 4 is $\dfrac{1}{3}.$

(iii) Let C be the event of getting a number less than 5, then

C = {1, 2, 3, 4}.

∴ The number of favourable outcomes to the event C = 4.

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{6} = \dfrac{2}{3} .$

Hence, the probability of getting a number less than 5 is $\dfrac{2}{3}.$

(iv) Let D be the event of getting the number 5, then

D = {5}.

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

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{6}.$

Hence, the probability of getting the number 5 is $\dfrac{1}{6}.$