Question

$20$ cards are numbered from $1$ to $20$. One card is drawn at random. What is the Probability that the number on the card is
(i) A multiple of $4$
(ii) Not a multiple of $6$

Answer 1

Given, $S=1,2,..........20$
$\therefore n\left(s\right)=20$
Let $A$ be event of getting multiple of $4$.
$\Rightarrow A=4,8,12,16,20$
$\Rightarrow n\left(A\right)=5$
Therefore, $P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{5}{20}=\frac{1}{4}$
Let $B$ be event of getting multiple of $6$.
$\Rightarrow B=6,12,18$
$\Rightarrow n\left(B\right)=3$
Therefore, $P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{3}{20}$
Probability of getting not a multiple of $6$.
$P\left(\overline{B}\right)=1-P\left(B\right)=1-\frac{3}{20}=\frac{17}{20}$