# Probability of Specific Distrubution

Lets say I have an event that has three possible outcomes (A, B, C) that are equally likely to occur. How do I calculate the probability of (for example) 3 A's, 2 B's, and 1 C when order doesn't matter? Specifically how do I calculate the number of permutations for any given distribution without a brute force approach that isn't feasible for a large number of trials?

E.g. for 4 trials

A, B, C
0, 0, 4 = 1
0, 1, 3 = 4
0, 2, 2 = 6
0, 3, 1 = 4
0, 4, 0 = 1
1, 0, 3 = 4
1, 1, 2 = 12
1, 2, 1 = 12
1, 3, 0 = 4
2, 0, 2 = 6
2, 1, 1 = 12
2, 2, 0 = 6
3, 0, 1 = 4
3, 1, 0 = 4
4, 0, 0 = 1

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You sample space doesn't make sense to me. What are your trials here? – jay-sun Nov 20 '12 at 17:30

First we describe the sample space. This consists of all sequences of length $6$ made up of letters chosen from our $3$-letter alphabet. There are $3^6$ such sequences, and by assumption they are all equally likely.

Now we have to count the "favourables," the sequences of length $6$ that have $3$ A's, $2$ B's, and (therefore) $1$ C.

The location of the A's can be chosen in $\dbinom{6}{3}$ ways. For each such way, there are $\dbinom{3}{2}$ ways to choose the location of the B's, for a total of $\dbinom{6}{3}\dbinom{3}{2}$.

Now divide by $3^6$.

Remark: Let us change the problem so that we still have independence, but we drop the requirement of equally likely. So suppose that A has probability $a$, B has probability $b$, and C has probability $c$, where $a+b+c=1$.

Any particular sequence made up of $3$ A's, $2$ B's, and $1$ C has probability $a^3b^2c^1$. There are $\dbinom{6}{3}\dbinom{3}{2}$ such sequences, so our probability is $$\binom{6}{3}\binom{3}{2}a^3b^2c^1.$$ In the case of your question, we have $a=b=c=\dfrac{1}{3}$.

The same basic strategy can be used for any similar problem. For example, if we modify your problem slightly to have $4$ equally likely outcomes A, B, C, D, then the probability of $4$ A's, $2$ B's, $1$ C and $3$ D's in $10$ trials is $\dbinom{10}{4}\dbinom{6}{2}\dbinom{4}{1}$ divided by $4^{10}$.

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