$A$ bag contains $24$ balls of which $x$ are red,$2x$ are white,and $3x$ are blue. $A$ ball is selected at random. What is the probability that it is:
$(i)$ not red?
$(ii)$ white?

(A) Given that,the total number of balls in the bag $= 24$.
Number of red balls $= x$,number of white balls $= 2x$,and number of blue balls $= 3x$.
According to the condition,$x + 2x + 3x = 24$.
$6x = 24$,which gives $x = 4$.
Therefore,the number of red balls $= 4$,the number of white balls $= 2 \times 4 = 8$,and the number of blue balls $= 3 \times 4 = 12$.
The total number of outcomes $n(S) = 24$.
$(i)$ Let $E_1$ be the event of selecting a ball which is not red. This means the ball can be white or blue.
$n(E_1) = \text{Number of white balls} + \text{Number of blue balls} = 8 + 12 = 20$.
Required probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{20}{24} = \frac{5}{6}$.
$(ii)$ Let $E_2$ be the event of selecting a ball which is white.
$n(E_2) = \text{Number of white balls} = 8$.
Required probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{8}{24} = \frac{1}{3}$.