An unbiased dice is thrown. What is the probability of getting

- (i) an even number
- (ii) a multiple of 3
- (iii) an even number or a multiple of 3
- (iv) an even number and a multiple of 3

**Solution**

In a single throw of an unbiased dice, you can get any one of the outcomes: 1, 2, 3, 4, 5, or 6.

So, exhaustive number of cases = 6.

(i) An even number is obtained if you obtain any one of 2, 4, 6 as an outcome.

So, favourable number of cases = 3.

Thus, required probability = 3/6 = 1/2

(ii) A multiple of 3 is obtained if you obtain any one of 3, 6 as an outcome.

So, favourable number of cases = 2

Thus, required probability = 2/6 = 1/3

(iii) An even number or a multiple of 3 is obtained in any of the following outcomes 2, 3, 4, 6.

So, favourable number of cases = 4.

Thus, required probability = 4/6 = 2/3

(iv) An even number and a multiple of 3 is obtained if you get 6 as an outcome.

So, favourable number of cases = 1.

Thus, required probability = 1/6