There are $2$ red marbles, $3$ white marbles and $5$ black marbles in a bag. What is the probability of drawing $1$ red marble, $2$ white marbles and $3$ black marbles, if $6$ marbles are drawn without replacement?

Answer: $$\frac{\binom{2}{1}\binom{3}{2}\binom{5}{3}}{\binom{10}{6}}=\frac{2}{7}$$

My question is why is the numerator of the answer $\binom{2}{1}\binom{3}{2}\binom{5}{3}$ instead of $1$? Below is my answer:

Total number of ways to choose $6$ marbles from $10$ marbles is $\binom{10}{6}$, so combinations such as

  • $BBBBBW$, $RWWBBB$, $WWWRBB$, ...

So the probabilty I calculated is $\frac{1}{\binom{10}{6}}$ because $RWWBBB$ is one of the possible combinations out of the $\binom{10}{6}$ total combinations. However this is incorrect, and is supposed to be $\frac{\binom{2}{1}\binom{3}{2}\binom{5}{3}}{\binom{10}{6}}$. I do not understand why the numerator is the product combinations because it does not make sense to multiply combinations of nondistinct objects. For example,

there are $5$ black marbles $BBBBB$ and so the number of ways to select $3$ black marbles is $1$, therefore it does not make sense to me that it is written as $\binom{5}{3}$ as above. To me this would make sense if it were asking to arrange $3$ letters at a time from $ABCDE$ where the order does not matter.

Can anyone explain why the numerator is the product of combinations instead of $1*1*1=1$?

My observation:

I noticed that using subscripts are necessary when there are fewer distinct objects than the total number of objects. See below:

The combination formula is $\binom{n}{r}$ where $n$ is total number of objects and $r$ is number of selected objects.

Let $k$ be the number of distinct objects where

  • if $k<n$, then subscripts are needed

  • if $k=n$, then subscripts are not needed

Case $1$. $k<n$

Suppose I want to select $r=2$ marbles from a bag with $1$ white marble and $2$ black marbles, so

  • $k=2$ and $n=3$, so $k<n$
  • So $\binom{n}{r}=3$ with combinations: {$W_1,B_1$}, {$W_1,B_2$}, {$B_1,B_2$}

Case $2$. $k=n$

Suppose I want to select $r=2$ marbles from a bag with $1$ white marble, $1$ black marble and $1$ red marble, so

  • $k=3$ and $n=3$, so $k=n$
  • So $\binom{n}{r}=3$ with combinations: {$W,B$}, {$W,R$}, {$B,R$}
  • $\begingroup$ @GrahamKemp I do not think the order matters in this question, so wouldn't $WWRBBB$, $RWWBBB$, $BBBWWR$,etc be the same, hence 1 combination? $\endgroup$
    – user314
    Feb 10, 2021 at 23:45
  • $\begingroup$ There are many ways to select RWWBBB, since we are making a choice of 1 from 2 red, 2 from 3 white, and 3 from 5 blue. It is not the order of colour, but the distinct marbles that matters. $\endgroup$ Feb 10, 2021 at 23:56

2 Answers 2


Consider a simpler problem: your bag has $99$ blue marbles and $1$ red marble in it. You draw a marble from the bag. What is the probability that it's blue?

If you want to answer $\frac{99}{100}$ rather than $\frac12$, you want to treat the blue marbles as distinguishable. This has nothing to do with whether you can tell the marbles apart by looking.

Rather, the problem is that we can only find probabilities by counting outcomes if we are sampling uniformly. In order to be drawing one of the $100$ marbles uniformly, the $100$ marbles should all be considered different outcomes.

Similarly, in the actual question you're dealing with, we can only answer the question by counting outcomes if all $10$ marbles are distinguishable: because only in that case are there $\binom{10}{6}$ different outcomes, each of which is equally likely. So you shouldn't be thinking of your outcomes as $BBBBBW, RWWBBB, \dots$. Rather, we should say:

  1. There are $10$ marbles in the bag: marbles $R_1, R_2$ are red, marbles $W_1, W_2, W_3$ are white, and marbles $B_1, B_2, B_3, B_4, B_5$ are black.
  2. We draw a uniformly random subset of $6$ of the marbles, such as $\{B_1, B_2, B_3, B_4, B_5, W_1\}$ or $\{R_2, W_1, W_3, B_2, B_4, B_5\}$.

In this case, there are in fact $\binom21 \binom32 \binom53$ outcomes that match the description "$1$ red marble, $2$ white marbles, and $3$ black marbles". There's $\{R_1, W_1, W_2, B_1, B_2, B_3\}$ and $\{R_1, W_1, W_2, B_1, B_2, B_4\}$, and so on.

These are exactly all outcomes of the form $\{R_a, W_b, W_c, B_d, B_e, B_f\}$ where $\{a\} \subset \{1,2\}$, $\{b,c\} \subset \{1,2,3\}$, and $\{d,e,f\} \subset \{1,2,3,4,5\}$. From this description, we can see why there's $\binom21 \binom32 \binom53$ of them: there's $\binom21$ ways to pick $\{a\}$, $\binom32$ ways to pick $\{b,c\}$, and $\binom53$ ways to pick $\{d,e,f\}$.

If instead our outcomes only had descriptions like "there are $3$ black marbles and $3$ white marbles" without distinguishing the marbles of each color, then our outcomes wouldn't all be equally likely. There are actually only $7$ different outcomes in this case: $$WWWBBB, WWBBBB, WBBBBB,\\ RWWWBB, RWWBBB, RWBBBB, RBBBBB.$$ But their probabilities are different, so this model doesn't lend itself to computation very well.

  • $\begingroup$ I understand everything now except the reason why it is necessary to add the subscripts for each colored ball. For example, {$B_1,B_2,B_3,B_4,W_1,W_3$} is distinct from {$B_2,B_1,B_3,B_4,W_1,W_2$}, but the result is the same: $4$ black balls are chosen and $2$ white balls is chosen. So why are the subscripts required? $\endgroup$
    – user314
    Feb 11, 2021 at 0:41
  • $\begingroup$ Selecting $6$ out of $10$ distinguishable marbles is uniform: all the outcomes are equally likely. Selecting $6$ marbles out of $10$ imperfectly distinguishable ones isn't uniform, which means we can no longer find probabilities by counting outcomes. So we artificially make all $10$ marbles distinguishable. $\endgroup$ Feb 11, 2021 at 0:46
  • $\begingroup$ I updated my answer and want to know if that is the reason why you label $B,B,B,...$ as $B_1,B_2,B_3,..$ $\endgroup$
    – user314
    Feb 11, 2021 at 1:44
  • $\begingroup$ Yes, you are right. $\endgroup$ Feb 11, 2021 at 3:02

$\binom{10}6$ is the count for selecting 6 items from a set of 10 distinct items. We don't care about the order in which these are selected, just which individual marbles are selected.

So too should we not be concerned with order of the favored event. We count it in the same sort of way --- treating the marbles as distinct items although grouped by colours.

Thus we evaluate the probability for obtaining: 1 from 2 red, 2 from 3 white, and 3 from 5 blue, when selecting any 6 from all 10 marbles as:$$\dfrac{\dbinom 21\dbinom 32\dbinom 53}{\dbinom{2+3+5}{1+2+3}}$$


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