There are $400$ televisions where exactly $16$ are defect. What is probability..? > There are in total $400$ televisions where exactly $16$ of these are defect. What's the probability that from $25$ televisions, at most $2$ are defect? This experiment is without replacement (this means if you take a television you don't put it back in the set again). This is other example they can ask in test I write next week.. Not know good solution for it. You have in total $400$ television and $16$ of them is defect. I calculate probability that $1$ of the $400$ is defect: $$\frac{16}{400}= \frac{1}{25}$$ Now how I understand task, we have $25$ television which is subset of the total $400$ televisions. We need know how many of the $25$ television can be defect and then use binomial formula on it? How many of $25$ television can be defect? You don't can know because maybe no is defect. Very confused.. : /

A probability space consists of $C^{25}_{400}$ objects. So $|\Omega| = C^{25}_{400}$.

16 tvs are defected, 384 are not.

Let $A$ be an event that satisfies condition in problem. What is its cardinal number?

$$ |A| = C^{24}_{384} \cdot C^{1}_{16} + C^{23}_{384}\cdot C^{2}_{16} + C^{22}_{384} \cdot C^3_{16} $$

We conclude that probability you want to find is $\frac{|A|}{|\Omega|}$.

It would be an exact answer. But for sure it cannot be calculated easily...

But an answer can be approximated if you use De Moivre–Laplace theorem.

$p$ here is $\frac{16}{400} = 0.04$, $n = 25$. $x_1 = \frac{\mu - np}{\sqrt{npq}} = \frac{0 - 1}{\sqrt{0.96}} \approx -1.0206 $, $x_2 = \frac{2 - 1}{\sqrt{0.96}} = 1.0206$

$$ P(A) \approx \Phi(x_2 - \frac 12) - \Phi(x_1 + \frac 12) \approx 0.4 $$