In a single throw of a die, find the probability of getting a number

(i) greater than 2

(ii) less than or equal to 2

(iii) not greater than 2.

(i) Total number of possible outcomes = 6 (1, 2, 3, 4, 5 and 6)

Number of favourable outcomes (Getting a number greater than 2) = 4 (3, 4, 5 and 6)

P(Getting a number greater than 2) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{4}{6}$

= $\dfrac{2}{3}$

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

(ii) Total number of possible outcomes = 6 (1, 2, 3, 4, 5 and 6)

Number of favourable outcomes (Getting a number less than or equal to 2) = 2 (1, 2)

P(Getting a number less than or equal to 2) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{2}{6}$

= $\dfrac{1}{3}$

Hence, the probability of getting a number less than or equal to 2 is $\dfrac{1}{3}$.

(iii) Total number of possible outcomes = 6 (1, 2, 3, 4, 5 and 6)

Number of favourable outcomes (Getting a number not greater than 2) = 2 (1, 2)

P(Getting a number not greater than 2) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{2}{6}$

= $\dfrac{1}{3}$

Hence, the probability of getting a number not greater than 2 is $\dfrac{1}{3}$.