Random Probability!

The probability that the number is a multiple of $3$ is $\frac { 10 }{ 30 }$ . ( since $3\times 10=30$ . ) Similarly the probability that the number is a multiple of $13$ is $\frac { 2 }{ 30 }$ . (since $13\times 2=26$ . ) Neither $3$ nor $13$ has a common multiple from $1$ to $30$ . Therefore chance that the selected number is a multiple of $3$ or $13$ is $\frac { 10+2 }{ 30 } =\frac { 2 }{ 5 }$ .

multiples of 3 are 10 in first 30 num whereas multiples of 13 are 2 in num ....there are no common multiples in first 30 num ...so answer will be 10/30 + 2/30=12/30=2/5

Looking at sizes, the set S of the first 30 natural numbers has a size |S| of 30. The outcome set X, is the multiples of 3 and of 13, so has a size |X| of $10 + 2 = 12$ (no common multiples). Thus $P(X) = \frac {|X|}{|S|} = \frac {12}{30} = \frac {2}{5}$

The probability for the numbers to be a multiple of 3 is 10/30.The probability that the number is a multiple of 13 is 2/30.their combined probability is (10/30)+(2/30)-(0/30)=12/30 which results 2/5 and answer fallows.