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$10$ 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.

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  • $\begingroup$ 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 $1/3$? $\endgroup$
    – sudeep5221
    Dec 12, 2023 at 22:52
  • $\begingroup$ Your posting is missing several details. Please see this article on MathSE protocol. $\endgroup$ Dec 12, 2023 at 23:19

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – dmk
    Dec 13, 2023 at 11:57
  • $\begingroup$ @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". $\endgroup$ Dec 13, 2023 at 23:38
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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 :).

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