Question

Thirty identical cards are marked with numbers $1$ to $30$. if one card is drawn at random, find the probability that it is:
(i) a multiple of $4$ or $6$
(ii) a multiple of $3$ and $5$
(iii) a multiple of $3$ or $5$

Answer 1

There are $30$ cards from which one card is drawn .

Total number of elementary events $=n\left(S\right)=30$

$\left(i\right)$ From numbers $1$ to $30$, there are $10$ numbers which are multiple of $4$ or $6$ i.e. $\{4,6,8,12,16,18,20,24,28,30\}$

Favorable number of events $=n\left(E\right)=10$

Probability of selecting a card with a multiple of $4$ or $6=\frac{n\left(E\right)}{n\left(S\right)}=\frac{10}{30}=\frac{1}{3}$

$\left(ii\right)$ From numbers $1$ to $30$, there are $2$ numbers which are multiple of $3$ and $5$ i.e. $\{15,30\}$

Favorable number of events $=n\left(E\right)=2$

Probability of selecting a card with a multiple of $3$ and $5=\frac{n\left(E\right)}{n\left(S\right)}=\frac{2}{30}=\frac{1}{15}$

$\left(iii\right)$ From numbers $1$ to $30$, there are $14$ numbers which are multiple of $3$ or $5$ i.e. $\{3,5,6,9,10,12,15,18,20,21,24,25,27,30\}$

Favorable number of events $=n\left(E\right)=14$

Probability of selecting a card with a multiple of $3$ or $5=\frac{n\left(E\right)}{n\left(S\right)}=\frac{14}{30}=\frac{7}{15}$