What can I do if $X \sim H (N, M, n)$ but $N$ is not specified?

Question

Context

$35$% of the students who took the first semester of the Computer Technologist, exonerated the subject MDyL1. It is considered a sample of $10$ students of that semester and the random variable $X$: number of students that have the subject MDyL1 exonerated, among the $10$ selected.

• a) Calculate the probability that more than two students in the sample have exempted the subject MDyL1.
• b) Find the probability that less than half of the students in the sample have exempted the subject MDyL1.
• c) Determine $E (X)$ and $Var (X)$.

$$X \sim H (N, M, n) \Rightarrow P(X=x)=\dfrac{C_x^M\times C_{n-x}^{N-M}}{C_n^N}$$

I suppose $N$ is $100$% but $n$ is not in percent. Then what can I do?

Thank you very much.

Answer 1

The question contains insufficient information to determine a unique solution. To see why, suppose there are only $N = 20$ students in total; that is to say, the sample of $n = 10$ students comprises half of the entire population. Then the number of students who have "exempted"/"exonerated" (the meaning has been lost in translation, it seems), is $M = (0.35)N = 7$, and the random number $X$ of such students in the sample is hypergeometric with probability mass function $$\Pr[X = x] = \frac{\binom{M}{x} \binom{N - M}{n-x}}{\binom{N}{n}} = \frac{\binom{7}{x}\binom{13}{10-x}}{\binom{20}{10}}$$ and the desired probability is $$1 - \Pr[X \le 2] = \frac{533}{646} \approx 0.825077.$$ But now suppose $N \gg 10$, say $N = 10^6$ students. Then the sample can be effectively regarded as occurring with replacement, and the desired probability is well approximated by a binomial distribution: $$1 - \Pr[X \le 2] \approx 1 - \sum_{x=0}^2 \binom{10}{x} (0.35)^x (0.65)^{10-x} = \frac{1890285078059}{2560000000000} \approx 0.738393.$$

In general, if $N$ is unknown, you cannot do much better than to write $$\Pr[X > 2] = 1 - \sum_{x=0}^2 \frac{\binom{7k}{x} \binom{13k}{10-x}}{\binom{20k}{10}}, \quad N = 20k,$$ since in order for there to be exactly $35\%$ of students in one category the total number of students must be an integer multiple of $20$.