Question

A box contains $4$ white balls, $6$ red ball, $7$ black balls and $3$ blue balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is :
(i) neither white nor black
(ii) red or white, and
(iii) either white or red or black or blue

Answer 1

Given, white ball $=4$
Red ball $=6$
Black ball $=7$
Blue ball $=3$
Total number of balls $=20$
Therefore, $n\left(S\right)$ $=20$
(i) Let $A$ be event of getting neither White nor Black ball.
$\Rightarrow n\left(A\right)=6+3=9$
$\Rightarrow P\left(A\right)$ $=\frac{n\left(A\right)}{n\left(S\right)}=\frac{9}{20}$
(ii) Let $B$ be event of getting Red or White ball.
$\Rightarrow n\left(B\right)=6+4=10$
$\Rightarrow P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{10}{20}=\frac{1}{2}$
(iii) Let $C$ be event of getting either White or Red or black or blue ball.
$\Rightarrow n\left(C\right)=4+6+7+3=20$
$\Rightarrow P\left(C\right)=\frac{n\left(c\right)}{n\left(s\right)}=\frac{20}{20}=1$