# 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.

• Think of it as a single draw of $3$ balls. It has the exact same "impact" as $3$ draws of a single ball, hence the probability distribution is the same in both cases. – barak manos Sep 15 '16 at 15:00
• @barakmanos how the impact can be same? We're picking up balls without replacement. So if I picked up three balls at once the probability of choosing certain number of red balls is different from that in the case of picking one by one – Shuvam Shah Sep 15 '16 at 15:04
• it seems that the book solution is with replacement – user354674 Sep 15 '16 at 15:06

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

$k=0:\dfrac{\binom{4}{0}\binom{5}{3}}{\binom{9}{3}}=\dfrac{10}{84}\doteq0.11904762$
$k=1:\dfrac{\binom{4}{1}\binom{5}{2}}{\binom{9}{3}}=\dfrac{40}{84}\doteq0.47619048$
$k=2:\dfrac{\binom{4}{2}\binom{5}{1}}{\binom{9}{3}}=\dfrac{30}{84}\doteq0.35714286$
$k=3:\dfrac{\binom{4}{3}\binom{5}{0}}{\binom{9}{3}}=\dfrac{4}{84}\doteq0.04761905$

• Sorry but I did not understand your answer. Can you please elaborate on how you get to that above function? – Shuvam Shah Sep 15 '16 at 15:22
• I have simplified the thinking, but if you would like, I can elaborate the pick by pick probability method, too. – robjohn Sep 15 '16 at 15:28

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 -- ;
}