A die is thrown. Find the probability of

We know that,

Probability of occurrence of an event

$=\frac{\text { Total no. of Desired outcomes }}{\text { Total no. of outcomes }}$

As 3,6 are multiples up to 6, so the desired outcomes are 3,6, and total outcomes are $1,2,3,4,5,6$

Therefore, total no. of outcomes are 6, and total no. of desired outcomes are 2

Probability of getting multiple of 3

$=\frac{2}{6}$

$=\frac{1}{3}$

Conclusion: Probability of getting multiple of 3 when die is thrown is $\frac{1}{3}$