# Multinomial probability calculation [closed]

$$10$$ cars choose uniformly at random between three parking lots ($$A$$, $$B$$, and $$C$$). Calculate the probability that none of the parking lots are empty and that there are exactly $$5$$ cars in lot $$A$$.

How would you calculate this probability? As far as I can tell, it requires the use of multinomial distributions.

• What have you tried so far? Where are you stuck? Also does "choose at random" imply "choose uniformly at random", i.e., probability of choose each parking lot is $1/3$? Dec 12, 2023 at 22:52

Here are two possible ways of computing this:

1. Simply calculate the probability of a single arrangment that satisfies the requirement of exactly 5 cars in $$A$$ (treat $$B$$ and $$C$$ as a single unit with a probability of $$2/3$$) then multiply the answer by all possible ways to choose 5 cars out of 10: $$10 \choose 5$$. Next calculate the probability of a single arrangment with 5 in $$A$$ and 5 in $$B$$ i.e. an empty lot: and again multiply the answer by $$10 \choose 5$$. Subtract this second answer from your first answer two times (for $$B$$ and $$C$$) and you should be good.

2. Use the Multinomial Theoroem and sum all the values of $$10! \over 5!B!C!$$ for all non zero values of $$B$$ and $$C$$ (there are only a handful) This gives us the number of arrangments that satisfy the requirements. Now just divide by the amount of total arrangments.

• It was the middle of the night when I was writing up my solution, and so I couldn't figure out how to formally express your first solution :). But yes, this is (essentially) $P(A=5,1\le B\le4) = P(1\le B\le 4\; |\; A = 5) P(A=5)$. $A$'s marginal distribution is binomial, with $p=1/3$ and $10$ trials, while the distribution of $B\; | \; A$ is binomial with $p = \frac{p_B}{p_B+p_C}$ and $10-A$ trials.
– dmk
Dec 13, 2023 at 11:57
• @dmk All good. Someone posted a wrong answer the other day which motivated me to post this answer to what is basically a "Please do my homework question". Dec 13, 2023 at 23:38

I'll assume here that when the cars enter "at random", they enter uniformly at random.

Let $$A, B, C$$ stand for the number of cars entering the relevant lots. Then the pmf of this multinomial distribution, as you most likely know, is

$$P(A = a, B = b, C = c) = \binom{10}{a, \; b, \; c} \; p_A^a \cdot p_B^b \cdot p_C^c,$$

where $$a + b + c = 10$$ and where $$p_A$$, for example, is the probability of a car entering lot A. Since the cars enter any of the three lots with equal probability, we can write $$p_A = p_B = p_C = p = 1/3$$, and so our pmf simplifies to

$$P(A = a, B = b, C = c) = \binom{10}{a, \; b, \; c} \cdot \frac{1}{3^{10}}.$$

Now, we're interested in when $$a = 5$$ and neither $$b$$ nor $$c$$ is $$0$$. Another way of phrasing the latter restriction is that $$1 \le b \le 4$$. Since $$a+b+c=10$$, and $$a = 5$$, we must have $$c = 5 - b$$. Hence,

$$P(A = 5, 1 \le B \le 4) = \sum_{b=1}^{4} \binom{10}{5, \; b, \; 5 - b} \cdot \frac{1}{3^{10}}.$$

From here, you'll be able to compute the answer :).