If one ticket is randomly selected from tickets numbered 1 to 30, then find the probability that the number on the ticket is (i) a multiple of 5 or 7 (ii) a multiple of 3 or 5.

I) One ticket is randomly selected from  $1\text{ to }30,n\left(S\right){=}^{30}{C}_1$

i) A be the event of select a number which is multiple of 5

$A=\{5,10,15,20,25,30\}\Rightarrow n\left(A\right)=6$

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{6}{30}$

B be the event of selecting a number which is multiple of 7

$$\begin{gathered} B=\{7,14,21,28\}\Rightarrow n\left(B\right)=4 \\ \Rightarrow P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{4}{30} \end{gathered}$$

$A\cap B$ is the event of both multiple of 5 and 7

$A\cap B=\varphi ,n(A\cap B)=0,P(A\cap B)=0$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=\frac{6}{30}+\frac{4}{30}-0=\frac{10}{30}=\frac{1}{3}$

II) $A$ be the event of selecting a no. which is multiple of 3

$n\left(A\right)=10P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{10}{30}$

$B$ be the event of selecting a no.which is multiple of 5

$n\left(B\right)=6P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{6}{30}$

a $A\cap B$ is the event of multiple of both $3 \& 5$ 

$(A\cap B)=\{15,30\},n(A\cap B)=2$

$P(A\cap B)=\frac{2}{30}$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=\frac{10}{30}+\frac{6}{30}-\frac{2}{30}=\frac{14}{30}=\frac{7}{15}$