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Here is a question that I want to understand how to solve it.

At a university, a lecture hall holds $n = 400$ students. Realizing that some students invariably drop a class, the registrar overbooks the lecture hall for a particular class, allowing $m = 402$ students to register. Suppose that the probability any particular student drops the class is $p = 0.01$.

What are the exact and approximate distributions of $X$, the number of students dropping the class?

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up vote 3 down vote accepted

Assume each student acts independently of others. Let $U_i$ be 0-1 random variable, equal to 1 is $i$-th student drops the class. We have $\mathbb{P}(U_i = 1) = p$ and $\mathbb{P}(U_i = 0) = 1 - p$, i.e. $U_i$ are independent identically distributed Bernoulli random variables.

The total number of students dropping the class is $X = U_1 + U_2 + \ldots$. The sum of $n$ independent identically distributed Bernoulli random variables is equal in distribution to a Binomial random variable, thus $X \sim \operatorname{Binom}(n,p)$.

Since the expected number of student dropping the class is low $\mathbb{E}(X) = n p = 4$, Poisson approximation is applicable, thus approximately $X \sim \operatorname{Poisson}(n p)$. Indeed, compare PDFs graphically:

enter image description here

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The number of students who drop has expected value $4$. The number of "trials" ($400$) is so large that a Poisson distribution with expected value $4$ could be used as a reasonable approximation. That means the probability that exactly $x$ students drop is $$ \frac{4^x e^{-4}}{4!} = \frac{4^x e^{-4}}{4\cdot3\cdot2\cdot1}. $$

The normal approximation doesn't make huge amount of sense when the expected number of successes is as small as $4$. But it's probably not too bad if you use a continuity correction.

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