12. Stunning news. 18 scientists have arrived at the conference, and exactly 10 of them know the stunning news. During the break (coffee break), all scientists randomly pair up, and in each pair, each scientist who knows the news tells it to the other if they haven't heard it yet.

a) (from 6th grade. 1 point). Find the probability that after the coffee break, the number of scientists who know the news will be 13.

b) (from 10th grade. 4 points). Find the probability that after the coffee break, the number of scientists who know the news will be 14.

c) (from 9th grade. 3 points). Let $X$ be the number of scientists who know the stunning news after the coffee break. Find the expected value of $X$.
Solution. a) After the coffee break, the news will be known by everyone who ended up in a pair where there is at least one "knower" (someone who knew the news beforehand). Since these scientists form an even number of pairs, their number is even. Therefore, the probability that there will be 13 of them is zero.

b) The problem resembles the problem 7 about tennis players, as it boils down to counting pairs consisting of a "knower" and a "non-knower." For simplicity, let's assume the scientists are seated at two-person tables. We will color the seats of the knowers gray and leave the seats of the non-knowers white.

There are $C_{18}^{10}=C_{18}^{8}=43758$ ways to seat 10 knowers and 8 non-knowers on 18 seats around two-person tables. We are interested in such seating arrangements where exactly 4 knowers end up at a table with a non-knower, so they can pass on their knowledge and increase the number of knowers to 14 (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_853fe0fb72821c0a3a48g-15.jpg?height=280&width=1174&top_left_y=488&top_left_x=515)

Let's reason as in problem 7: choose 4 out of 9 tables for pairs of "knower-non-knower" (these tables are covered with gray tablecloths in the figure). From the remaining 5 tables, choose 2 tables for non-knowers (white tablecloths). Note that pairs of "knower-non-knower" can be seated in $2^{4}$ ways, by swapping the knower and non-knower at each table. The figure shows one of these possible seating arrangements, and the total number of required seating arrangements is $C_{9}^{4} \cdot C_{5}^{2} \cdot 2^{4}=126 \cdot 10 \cdot 16=20160$.

Thus, the probability that the number of knowers will become exactly 14 is

$$
\frac{C_{9}^{4} \cdot C_{5}^{2} \cdot 2^{4}}{C_{18}^{8}}=\frac{20160}{43758}=\frac{1120}{2431} \approx 0.461
$$

c) We can solve the problem in part b) in a general form, thereby obtaining the distribution of the random variable $X$, and then find the expected value $\mathrm{E} X$. However, we will take a simpler approach by using indicators. We will number the non-knowers from 1 to 8 and introduce indicators $I_{k}$ for the number of non-knowers. $I_{k}=1$ if the non-knower number $k$ becomes a knower during the coffee break, and $I_{k}=0$ if they do not learn the news. Clearly,

$$
X=10+I_{1}+I_{2}+\ldots+I_{8}
$$

The distributions of all indicators are the same:

$$
I_{k} \sim\left(\begin{array}{cc}
0 & 1 \\
\frac{7}{17} & \frac{10}{17}
\end{array}\right)
$$

since in a pair with any non-knower, any of the remaining 17 scientists can sit, and exactly 10 of them are knowers. Therefore, $E I_{k}=\frac{10}{17}$. Then

$$
\mathrm{E} X=10+\mathrm{E} I_{1}+\mathrm{E} I_{2}+\ldots+\mathrm{E} I_{8}=10+8 \cdot \frac{10}{17}=14 \frac{12}{17} \approx 14.7
$$

Answer: a) 0; b) approximately 0.461; c) approximately 14.7.