**Question:**

A die is thrown. Find the probability of getting:

(i) a prime number

(ii) 2 or 4

(iii) a multiple of 2 or 3

**Solution:**

When a die is thrown, the possible outcomes are $1,2,3,4,5$ and 6 .

Thus, the sample space will be as follows:

$\mathrm{S}=\{1,2,3,4,5,6\}$

(i) Let $A$ be the event of getting a prime number.

There are 3 prime numbers $(2,3$ and 5$)$ in the sample space.

Thus, the number of favourable outcomes is 3 .

Hence, the probability of getting a prime number is as follows:

$\mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{6}=\frac{1}{2}$

(ii) Let $A$ be the event of getting a two or four.

Two or four occurs once in a single roll.

Therefore, the total number of favourable outcomes is 2 .

Hence, the probability of getting 2 or 4 is as follows:

$\mathrm{P}(\mathrm{A})=\frac{2}{6}=\frac{1}{3}$

(iii) Let A be the event of getting multiples of 2 or 3 .

Here, the multiples of 2 are $2,4,6$ and the multiples of 3 are 3 and 6 .

Therefore, the favourable outcomes are 2, 3,4 and 6 .

Hence, the probability of getting a multiple of 2 or 3 is as follows:

$P(A)=\frac{\text { Number of favorable outcomes }}{\text { Total number of outcomes }}=\frac{4}{6}=\frac{2}{3}$