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Imagine there are 'k' boxes. All boxes contain 'n' unique colored marbles. Assuming that we select 'm' marbles out of 'm' different boxes (one marble from each box), what is the probability that all of the selected marbles have the same color? (e.g. imagine we have five boxes (k = 5), in each box there is a red, blue, purple, and yellow marble (n = 4). Now we select 3 marbles from 3 different boxes (m = 3). What is the probability that all the 3 selected marbles have the same color)?

My solution:

The total possible ways of choosing 'm' marbles out of 'n' unique colored marbles from 'k' different boxes is:

A = ${n \choose m}\times{n \choose k}$

The desirable event can happen this many times:

B = $\left(\frac{1}{n^m}\right)$

So the answer will be: ${B\over A}$

I am not sure tho, so please correct me if I am wrong!

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In fact, it is as good as choosing a marble each out of $m$ boxes regardless of $k$.

Hence, the probability you are looking for is simply $$\binom n1 \left(\frac1n\right)^m=n^{1-m}$$


Or to follow your thought process, your $A$ should have been

$$A=\binom km \cdot n^m $$

Then for $B$, you have to consider

  • the number of ways the $m$ boxes are chosen out of $k$,

  • the number of ways the first color may be chosen, and

  • the number of ways the subsequent $(m-1)$ balls are of the same color

So your $B$ is in fact $$B=\binom km \binom n1 (1)^{m-1}$$

So your probability similarly becomes $$\frac BA=n^{1-m}$$ Notice that the factor $\binom km$ vanishes. Hence, it may in fact be ignored from the start.

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  • $\begingroup$ Thanks, but could you explain a little bit, why does the term $\binom km$ vanish?! I can see it mathematically, but in reality, why doesn’t it matter how we select ‘m’ out of ‘k’ boxes? Is it because the boxes are exactly the same (the same bc they contain the exact same marbles?)? $\endgroup$ – Antonio May 4 '18 at 3:53
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    $\begingroup$ @Antonio Yes, there are two reasons: 1. The boxes are all identical. It doesn't matter which boxes are chosen; 2. The very same boxes are chosen at the very start of both event and sample spaces. $\endgroup$ – Karn Watcharasupat May 4 '18 at 11:48

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