Say you have an urn with $2$ red balls, $2$ white balls, and $2$ black balls.

Draw $3$ balls from this urn.

What is the probability that all balls have a different colour?

I am going to answer this question in 4 different ways, and arrive at 3 different answers:

  1. We need not need to care about the order in which the balls are drawn to answer our question. Thus, the sample space (the space of all possibilities) is: $\Omega_1 = \{\text{rrw}, \text{rrb}, \text{wwr}, \text{wwb}, \text{bbr}, \text{bbw}, \text{rbw}\}$. The event corresponding to the case where all balls have a different colour is $E_1 = \{\text{rbw}\}$. The probability of this event occurring, i.e. the relative size of this event, compared to the entire sample space is trivial to obtain: $E_1/\Omega_1 = 1/7$.

  2. Say that we did care consider the order in which balls are drawn when determining the answer. Each ball is labeled with two indices: the first is the label of the ball within its own colour group, and the second is the draw order of the ball. So the event $\{\text{r}_{1,2}, \text{r}_{2, 2}, \text{w}_{2, 3}\}$ corresponds to the case where red ball $1$ is drawn first, followed by red ball $2$ second, followed by white ball $2$ third. Let us determine the size of $\Omega_2$ based on the elements of $\Omega_1$: each element in $\Omega_1$ with at least one off-colour ball corresponds to $12$ elements in $\Omega_2$ because we have $6$ choices for the order in which the $3$ balls are selected, and $2$ choices for which off-colour ball is selected. The last element in $\Omega_1$, $\text{rbw}$, corresponds to $8 \times 6 = 48$ elements in $\Omega_2$, since I have $2$ choices for each of the $3$ ball colours ($2 \times 2 \times 2 = 8$), and $6$ ways to order the selected balls. Note that $|E_2|$, the size of the event where all balls have a different colour, is thus $48$. So, $|\Omega_2| = 6 \times 12 + 6 \times 8 = 6 \times 20 = 120$, and $|E_2|/|\Omega_2| = 48/120 = 8/20 = 2/5$.

  3. Say that we cared about the order in which the balls are drawn without distinguishing between balls of the same colour when determining the answer. Let the sample space this time be denoted $\Omega_3$. Note that in $\Omega_1$, there are 6 cases where there is at least one off-colour ball, and each one of these corresponds to $3$ different cases in $\Omega_3$, since there are 3 different ways we can place the off-colour ball. The element where all 3 ball colours are different in $\Omega_1$ corresponds $6$ elements in $\Omega_3$, because there are $6$ ways to order 3 distinct objects. Thus, $|E_3| = 6$, and $|\Omega_3| = 6\times 3 + 1\times 6 = 24$. In particular, $|E_3|/|\Omega_3| = 6/24 = 1/4$.

  4. Let us calculate the probability that we draw 3 balls of different colour using a conditional probability approach. Draw the first ball. Now there are $5$ balls left in the urn, and $4/5$ of the balls are of a different colour than the first ball. Draw the second ball from this pool of $4$ balls. Now there are $4$ balls left, and $2$ are balls of a different colour from those already drawn, so there is a $2/4$ chance we draw a ball of yet another colour. Thus, the probability of drawing $3$ balls of different colours is $4/5 \times 2/4 = 2/5$.

I am doing something wrong, but I am not sure what. In particular, why is answer 3 different from answer 4? Second, why are answers 1, 2 and 3 different? Have I simply not restated the same problem in 4 different ways?

  • $\begingroup$ In $1$, your different elements in the sample space do not have equal probability. e.g. $P(rrw)=\frac1{5}$, and $P(brw)=\frac25$ which you can get by computing with brute force. In general, when computing sizes of sets, its crucial to make sure each element has the same probability, or it won't work. $\endgroup$ – John Doe Nov 4 '17 at 1:06

The first approach is wrong because the events in your sample space do not have the same probability. This is actually clarified in your second approach, where you show that each item in $\Omega_1$ can be realized by different numbers of outcomes.

Your second approach is correct because each item of $\Omega_2$ has equal probability.

I suspect your third approach is incorrect because you again do not have equal-probability events in $\Omega_3$. You seem to treat all color combinations as equally likely, when it is harder to end up with multiple balls of the same color. (Once you draw a red ball, it is harder to draw another red ball afterward.)

The fourth approach is fine; the reasoning via conditional probability is performed correctly.

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There is yet another way to get the probability: label the two balls of each color differently (as in your method 2) so that each of the six balls is uniquely identified and you can distinguish exactly which three of them were selected, but do not consider the order in which the three balls were drawn. There are $\binom 63 = 20$ possible ways to choose three items out of six distinguishable items without regard to the order of choosing. Since all six balls are equally uniquely identified, by symmetry all $20$ possible outcomes are equally likely.

Of those $20$ outcomes, the ones that have one ball of each color are the ones with either the first or second red ball, either the first or second white ball, and either the first or second black ball. There are $2\times 2\times 2 = 8$ of these combinations. Hence the probability is $8/20 = 2/5.$

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