professor has a coffee break when he reads the first unsuccessful exam. What is the probability that he has to read at least ($\geq$) 20 exams?

Question

In general, a student fails exam X with probability 0.2. At the end of today's exam, professor reviews each exam one by one, and he has a coffee break when he reads the first unsuccessful exam. What is the probability that he has to read at least ($\geq$) 20 exams before having a coffee break? Let's suppose number of students can be considered as infinite, and each exam is independent.

I can consider a Bernoulli distribution with probability 0.2, because each exam is independent. Because of this, I can use binomial distribution with a high number of students (let's suppose there are 1000 students). Now, I've used De Moivre-Laplace theorem to approximate binomial to a normal distribution. $$Z = N(np, np(1-p))$$ and so I get $$P(Z\geq 20) = P\left(N(0,1) \geq \frac{20.5\cdot np}{\sqrt{npq}}\right)$$

but when I do calculations I get incorrect results. $20.5\cdot np = 4100$, and that result is absurd. where's the problem? because by definition, when I have independent trials I can use bernoulli distribution, and binomial distribution, and De Moivre-Laplace is used when I have a lot of bernoulli's trials.

Answer 1

Isn't the task much easier?

The probability that the break will be after $n$-th student: $0.8^{n-1}\cdot 0.2$.

Answer 2

Trying to answer the OP's question "where's the problem"?

The normal distribution proposed by OP would approximate the number ($Z$) of failing exams in the entire batch of $1000$ (rather than when the first such exam would occur). The expected number of failed exams in the batch would be $200$, with a standard deviation of $\sqrt{160}$. The probability $P(Z\ge 20)$ is asking about the probability that there are at least $20$ failing exams (rather than: when will the professor first see a failing exam). As one might expect, $P(Z\ge 20)$ is extremely close to $1$ (coming from a normal distribution with mean $200$ and standard deviation $\sqrt{160}\approx 12.6$).

Also: The normal approximation gives $P(Z\ge 20)\approx P\left(N(0,1)\ge \frac{19.5-200}{\sqrt{160}}\right)\approx P\left(N(0,1)\ge -14.3\right)$. This probability value is essentially $1$.