14. Concept. In Anchuria, there are $K$ laws and $N$ ministers. The probability that a randomly chosen minister knows a randomly chosen law is $p$. One day, the ministers gathered for a council to write the Concept. If at least one minister knows a law, then this law will be considered in the Concept; otherwise, this law will not be considered in the Concept. Find:

a) The probability that exactly $M$ laws will be considered in the Concept.

b) The expected number of laws considered in the Concept.
Solution. a) Let's find the probability that no minister knows the first law: $(1-p)^{N}$. Therefore, the probability that the first law will be taken into account is $q=1-(1-p)^{N}$.

The probability that exactly $M$ of the $K$ laws will be taken into account in the Concept can be easily found using the binomial probability formula:

$$
\mathrm{P}_{M}=C_{K}^{M} q^{M}(1-q)^{K-M}=C_{K}^{M}\left(1-(1-p)^{N}\right)^{M}(1-p)^{N \cdot(K-M)}
$$

b) Let's introduce an indicator - a random variable $I_{k}$, which equals 1 if the law numbered $k$ is taken into account and 0 otherwise. Such a variable is called an indicator. The expected value is

$$
\mathrm{E} I_{k}=1 \cdot\left(1-(1-p)^{N}\right)+0 \cdot(1-p)^{N}=1-(1-p)^{N}
$$

The number of laws taken into account $X$ is equal to the sum of these variables: $X=I_{1}+I_{2}+\ldots+I_{K}$.

Therefore, $\mathrm{E} X=\mathrm{E} I_{1}+\mathrm{E} I_{2}+\ldots+\mathrm{E} I_{K}=K \cdot\left(1-(1-p)^{N}\right)$.

Answer: a) $C_{K}^{M}\left(1-(1-p)^{N}\right)^{M}(1-p)^{N \cdot(K-M)} ; \boldsymbol{\text { b) }} K\left(1-(1-p)^{N}\right)$.