In a single throw of a die, find the probability that the number :

(i) will be an even number

(ii) will be an odd number

(iii) will not be an even number.

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

Number of favourable outcomes (Getting an even number) = 3 (2, 4, 6)

P(Getting an even number) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{3}{6}$

= $\dfrac{1}{2}$

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

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

Number of favourable outcomes (Getting an odd number) = 3 (1, 3, 5)

P(Getting an odd number) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{3}{6}$

= $\dfrac{1}{2}$

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

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

Number of favourable outcomes (Getting not an even number) = 3 (1, 3, 5)

P(Getting not an even number) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{3}{6}$

= $\dfrac{1}{2}$

Hence, the probability of getting not an even number is $\dfrac{1}{2}$.