If one ticket is randomly selected from tickets numbered 1 to 30 then the probability that the number on the ticket is a multiple of 5 or 7

  $n\left(S\right)={30}_{C_1}$ 

$n\left(E_1\right)=6_{C_1} \left(multiple of 5\right)$

$n\left(E_2\right)=4_{C_1}, \left(multiple of 7\right)$         and  $n(E_1\cap E_2)=0$

$P(E_1\cup E_2)=P\left(E_1\right)+P\left(E_2\right)$

                        $=\frac{10}{30}=\frac{1}{3}$    

The probability that the number on the ticket is a multiple of 5 or 7 is $\frac{1}{3}$