Combinations of indistinguishable marbles

Question

Let's consider this problem:

A bag contains 5 black marbles and 6 white ones. Marbles of the same color are indistinguishable from each other. If I draw two marbles, what is the probability they have different colors?

This problem can be trivial if we reason like this:

$P(first = black, second=white) = P(first = black)*P(second=white|first = black) = \frac{5}{11} * \frac{6}{10} = \frac{3}{11}$

$P(first = white, second=black) = P(first = white)*P(second=black|first = white) = \frac{6}{11} * \frac{5}{10} = \frac{3}{11}$

Hence summing the two $P(\text{both black and white}) = \frac{6}{11}$.

If we frame the problem using combinations we can get the same result using this:

$P(\text{both black and white}) = \frac{\binom{6}{1}\binom{5}{1}}{\binom{11}{2}} = \frac{6}{11}$

This is like the number of combinations of choosing 1 marble from the white, 1 marble from the blacks and divide then by the number of combinations of choosing 2 marbles from the 11.

Is it a coincidence or is it correct this way of solving it? I cannot figure out how I should be allowed to use $\binom{11}{2}$ at denominator since to me there are at most 3 different ways to get a group of two marbles from the 11 namely (BB, BW, WW) and not 55 like $\binom{11}{2}$ states.

Please clarify this.

Answer 1

I cannot figure out how I should be allowed to use $11 \choose 2$ at denominator since to me there are at most $3$ different ways to get a group of two marbles from the $11$ namely (BB, BW, WW) and not $55$ like $11 \choose 2$ states.

Well, if you think about it this way, then you should feel uncomfortable with the numerator too. I mean there are only two ways to get $1$ black and $1$ white marble - BW or WB. So why ${6 \choose 1} \times {5 \choose 1}$? And if you take this line of thinking to its conclusion, the probability would be $2/3$ (incorrect, of course). So why $6/11$?

Here's why.

When it comes to probability, "indistinguishability" becomes a tricky concept. The question here states that marbles of the same color are indistinguishable. Does that mean there's only $1$ way to end up with a white marble on the first draw (W) and similarly one to end up with a black marble (B)? Does that mean the probability of drawing a black marble is the same as that of drawing a white one - assuming we draw only one?

The answer to the first question is - yes and no. And that to the second one is a clear no. Let me explain.

Imagine putting your hand inside the bag to draw a marble. Are there only $2$ marbles inside (because there are only two colors and marbles of the same color are indistinguishable)? No, there are $11$ marbles and you can draw any of them. So there are 11 different ways of drawing a marble. $5$ of these ways will give you a black marble and $6$ a white marble?

Yes, the end result of using any of the $5$ ways to draw a black marble would be indistinguishable. You end up with a black marble. But because the number of ways to draw a black marble is $5$ ways compared to the $6$ for a white marble, it's slightly less likely you will draw a black marble. To be precise, $P(B) = 5/11$, whereas $P(W) = 6/11$.

The point is - just by virtue of their separate existence, the marbles are distinguishable. When the question says they are indistinguishable, it's in the context that we don't care which black/white marble is drawn. So if it helps, you can reframe the question in your head as follows.

There are $5$ black marbles and $6$ white marbles. The marbles are all distinct. If I draw two marbles, what is the probability they have different colors?

Think about it, why should the answer to this question be any different from that to the original?

Answer 2

Both of your solutions are correct.

While marbles of the same color may be indistinguishable, we can treat them as distinguishable when calculating the probability that they have different colors. For instance, we could label the black marbles $B_1, B_2, B_3, B_4, B_5$ and the white marbles $W_1, W_2, W_3, W_4, W_5, W_6$. We do not care which of the five black marbles we select or which of the six white marbles we select. What matters is that a success is defined as selecting one of the five black marbles and one of the six white marbles from the bag when selecting two marbles from the bag without replacement. If we consider the order of selection, we obtain your first solution. If we do not consider the order of selection, we obtain your second solution.

Addendum: We treat the marbles as if they were distinguishable so that each marble is equally likely to be chosen. Imagine that the bag contains $999$ white marbles but only one black marble. Clearly, the probability of selecting the only black marble on the first draw is $1/1000$, not $1/2$. What matters is how many marbles of each color are in the bag, not that there are two colors of marbles in the bag.