Multinomial probability calculation

Question

$10$ 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.

Answer 1

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.

Answer 2

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 :).