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I have a problem with the following exercise. Given an urn with $w$ white balls, $r$ red balls and $b$ black balls, so in total there are $w+r+b$ balls. Now $3$ balls are drawn from the urn with replacement. What is the probability to get each colour.

I tried to solve it this way: It is equally likely to draw each ball (not taking care of the colour). The number of different ways to draw balls with replacement is described by combinations with replacement, say $C_{n,k}^{(r)} = \binom{n+k-1}{k}$. The probability should be therefore: \begin{align*} \frac{C_{w,1}^{(r)} C_{r,1}^{(r)} C_{b,1}^{(r)}}{C_{w+r+b,3}^{(r)}} = \frac{w r b}{\binom{w+r+b+2}{3}}. \end{align*} As I am highly sceptical about such exercises I wrote a short script testing the probability. The result is that either my mathematical approach or my short script is wrong. As I am pretty sure that my mathematical approach is wrong I would like to know why. Why can't I get the probability with this method?

Kind regards!

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The probability of drawing one of each colour in three draws with replacement is actually:$$\dfrac{3! r wb}{(r+w+b)^3}$$

Your formula simplifies to: $$\dfrac{3!rwb}{(r+w+b+2)(r+w+b+1)(r+w+b)}$$

The reason this is not the same is that $\mathrm C^{(\mathrm r)}_{n,k}$ counts the number of distinct combinations of $k$ items extracted from a set of $n$ items without distinguishing by order.   These are not equally probable outcomes.   So taking the ratio of ways to select combinations with replacement for one from each colour, against ways to select combinations with replacement for three from all balls, will not yield a valid probability measure.

For instance $\mathrm C^{\rm (r)}_{3,2}=6$ counts the ways to extract two balls from three, $\color{red}\bullet\color{red}\bullet, \color{red}\bullet\color{blue}\bullet, \color{red}\bullet\color{green}\bullet, \color{blue}\bullet\color{blue}\bullet, \color{blue}\bullet\color{green}\bullet, \color{green}\bullet\color{green}\bullet.$   Where as there would be $3^3$ equally probable ways: $\color{red}\bullet\color{red}\bullet, \color{red}\bullet\color{blue}\bullet, \color{red}\bullet\color{green}\bullet, \color{blue}\bullet\color{red}\bullet, \color{blue}\bullet\color{blue}\bullet, \color{blue}\bullet\color{green}\bullet, \color{green}\bullet\color{red}\bullet, \color{green}\bullet\color{blue}\bullet, \color{green}\bullet\color{green}\bullet.$

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You have these valid combination to get all three colors $(w,b,r),(w,r,b),(r,w,b),(r,b,w),(b,r,w),(b,w,r)$

for the first case : we need to get a white color then black then red. so the probability is $p_1 = \frac{w}{w+b+r} * \frac{b}{w+b+r-1} * \frac{r}{w+b+r-2} = \frac{6 b r w}{\binom{b+r+w }{3}}$

you will do the same for $p_2,p_3,p_4,p_5,p_6$ and then adding them together will result $6 p_1 = \frac{36 b r w}{\binom{b+r+w }{3}}$

hope its what you are looking for

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  • $\begingroup$ This looks very good, thank you very much for your answer! However, I would love to know why my approach does not work to fully understand this concept. $\endgroup$
    – blablablup
    Mar 18 '17 at 12:42

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