Picking certain number of balls without replacement and finding its probability distribution Suppose I am to choose three balls without replacement from a bag containing $5$ white and $4$ red balls. What will be the probability distribution of the red balls drawn ?.
According to my book, probability function will be
$$
{3\choose x}\left(\,{4 \over 9}\,\right)^{x}\left(\,{5 \over 9}\,\right)^{3 - x}
$$
What I didn't understand is why my book is taking probability of choosing red ball to be $4/9$ and the probability of choosing a white ball to be $5/9$. I think the above probabilities are of choosing the red and the white balls in the first trial. In other trials the probability of the above two events will change as we are drawing balls without replacement.
 A: Simplest Approach
Without replacement, the simplest method is to compute how many ways there are to pick $k$ from the $4$ red balls and $3-k$ from the $5$ white balls. Then divide that by the total number of ways to pick $3$ from the $9$ balls in total:
$$
\frac{\binom{4}{k}\binom{5}{3-k}}{\binom{9}{3}}
$$

Pick by Pick Approach
Without replacement, there are $\binom{3}{k}$ orders in which to draw $k$ red balls and $3-k$ white balls. Suppose we draw all the red balls first, the probability of that draw would be
$$
\underbrace{\overbrace{\frac{4}{9}\cdot\frac{3}{8}\cdots}}^{\text{drawing $k$ red balls}}_{k\text{ terms}}
\quad\underbrace{\overbrace{\frac{5}{9-k}\cdot\frac{4}{8-k}\cdots}}^{\text{drawing $3-k$ white balls}}_{3-k\text{ terms}}
=\frac{\frac{4!}{(4-k)!}\frac{5!}{(5-(3-k))!}}{\frac{9!}{(9-3)!}}
=\frac{\binom{4}{k}k!\binom{5}{3-k}(3-k)!}{\binom{9}{3}3!}
=\frac{\binom{4}{k}\binom{5}{3-k}}{\binom{9}{3}\binom{3}{k}}
$$
But no matter which order we drew the balls, the probability would be the same, the numerators would just change order. Thus, we just multiply the probability of each order by the number of orders:
$$
\frac{\binom{4}{k}\binom{5}{3-k}}{\binom{9}{3}\binom{3}{k}}\binom{3}{k}
=\frac{\binom{4}{k}\binom{5}{3-k}}{\binom{9}{3}}
$$

Values
$k=0:\dfrac{\binom{4}{0}\binom{5}{3}}{\binom{9}{3}}=\dfrac{10}{84}\doteq0.11904762$
$k=1:\dfrac{\binom{4}{1}\binom{5}{2}}{\binom{9}{3}}=\dfrac{40}{84}\doteq0.47619048$
$k=2:\dfrac{\binom{4}{2}\binom{5}{1}}{\binom{9}{3}}=\dfrac{30}{84}\doteq0.35714286$
$k=3:\dfrac{\binom{4}{3}\binom{5}{0}}{\binom{9}{3}}=\dfrac{4}{84}\doteq0.04761905$  
A: a simple simulation :
With replacement:


*

*0 red :17052 , /100000 = 0.1705 , book = 0.1714

*1 red :41091 , /100000 = 0.4109 , book = 0.4115

*2 red :33134 , /100000 = 0.3313 , book = 0.3292


Without replacement:


*

*0 red :11966 , /100000 = 0.1196 , book = 0.1714 , RobJohn = 0.119

*1 red :47916 , /100000 = 0.4791 , book = 0.4115 , RobJohn = 0.4761

*2 red :35484 , /100000 = 0.3548 , book = 0.3292 , RobJohn = 0.3571


book : ${3\choose x} \times (\frac{4}{9})^x \times (\frac{5}{9})^{3-x}$
RobJohn : $\frac{\binom{4}{k}\binom{5}{3-k}}{\binom{9}{3}}$
Then the book has a typo error on the word "without".
note : I treated the replacement basically, removing the ball from the datas , see the variables remain and remainred and how they are treated with the argument replace. Then it is credible.
for( j = 0 , red=0,remain=9 , remainred = 4; j < 3 ; j++ )
{
    dice = Math.floor(remain*Math.random()) ;
    if( dice < remainred ) // 0 base index
    {
        red ++ ; 
        if( ! replace ) 
            remainred -- ;
    }
    if( ! replace ) 
        remain -- ;
}

