If three balls are drawn one by one without replacement from a bag containing 5 white and 4 red balls, then the probability distribution of the number of white balls drawn is

The correct answer is Option (2) → 

Total balls $=9$ with $5$ white and $4$ red.

Total ways to choose any $3$ balls:

$\frac{9!}{3!6!}=84$

Let $X$ be the number of white balls drawn.

$P(X=0)=\frac{\frac{4!}{3!1!}}{84}=\frac{4}{84}=\frac{1}{21}$

$P(X=1)=\frac{\frac{5!}{1!4!}\cdot\frac{4!}{2!2!}}{84}=\frac{5\cdot6}{84}=\frac{30}{84}=\frac{5}{14}$

$P(X=2)=\frac{\frac{5!}{2!3!}\cdot\frac{4!}{1!3!}}{84}=\frac{10\cdot4}{84}=\frac{40}{84}=\frac{10}{21}$

$P(X=3)=\frac{\frac{5!}{3!2!}}{84}=\frac{10}{84}=\frac{5}{42}$

final answer: $P(0)=\frac{1}{21},\;P(1)=\frac{5}{14},\;P(2)=\frac{10}{21},\;P(3)=\frac{5}{42}$