A die is thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3
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:
$P(A)=\frac{\text { Number of favorable outcomes }}{\text { Total number of outcomes }}=\frac{4}{6}=\frac{2}{3}$