The probability of any event will be easier by using the standard formula which is mentioned in the question. An unbiased die is thrown, what is the probability of getting

${\text{(i)}}{\text{.}}$an even number or a multiple of 3

${\text{(ii)}}{\text{.}}$a number 3 or 4

${\text{(iii)}}{\text{.}}$a number greater than 3

Hint: In a single throw of dice, we will get 1, 2, 3, 4, 5 or 6, as the possible outcomes. No. of total outcomes$ = 6$. Use the standard formula, P(event)= (No. of favourable outcomes)/ (Total no. of possible outcomes).

Given, an unbiased die is thrown, so the sample space, $S = \{ 1,2,3,4,5,6\} $.

Therefore, the total no. of outcomes $ = 6$.

${\text{(i)}}{\text{.}}$an even number or a multiple of 3 are 2, 3, 4, 6

Therefore favourable no. of outcomes $ = 4$

 Using the formula, P(event)= (No. of favourable outcomes)/ (Total no. of possible outcomes)

to find the probability of an event.

So, $P$(even number or multiples of 3)$ = \dfrac{4}{6} = \dfrac{2}{3}$

${\text{(ii)}}{\text{.}}$a number 3 or 4

The number of favourable outcomes$ = 2$

$P$(a number 3 or 4) $ = \dfrac{2}{6} = \dfrac{1}{3}$

${\text{(iii)}}{\text{.}}$number greater than 3

The favourable outcomes are 4, 5, 6.

So, no. of favourable outcomes$ = 3$

$P$(a no. greater than 3) $ = \dfrac{3}{6} = \dfrac{1}{2}$

Note: When solving this type of questions, always think about all the favourable outcomes possible for a given event and the total outcomes possible. Then, if you know these two correctly then finding the probability of any event will be easier by using the standard formula which is mentioned in the question.