At a conference, 18 scientists arrived, exactly 10 of whom know a stunning piece of 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 the other hasn't heard it yet.

a) Find the probability that after the coffee break, the number of scientists who know the news will be 13.

b) Find the probability that after the coffee break, the number of scientists who know the news will be 14.

c) Let $X$ be the number of scientists who know the stunning news after the coffee break.

Find the expected value of $X$.
a) After the coffee break, the news will be known by everyone who ended up in a pair with 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 problem 65771, 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 chairs occupied by knowers gray and leave the chairs of non-knowers white.

There are $C_{18}^{10}=C_{18}^{8}=43758$ ways to seat 10 knowers and 8 non-knowers on 18 chairs around two-person tables. We are interested in seating arrangements where exactly four 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_08_22585010b2534c7ab9ebg-19.jpg?height=280&width=1158&top_left_y=89&top_left_x=447)

Choose four out of nine tables for the knower-non-knower pairs (on the figure, these tables are covered with gray tablecloths). From the remaining five tables, choose two tables for the non-knowers (white tablecloths). Note that the knower-non-knower pairs 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 desired 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) Number the non-knowers from 1 to 8 in some way and introduce indicators $I_{k}$ for the number of non-knowers: $I_{k}=$ 1 if 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:

$$
\left(\begin{array}{cc}
0 & 1 \\
\frac{7}{17} & \frac{10}{17}
\end{array}\right), \text { since any non-knower can be paired with }
$$

any of the remaining 17 scientists, and exactly 10 of them are knowers. Therefore, $\mathrm{E}_{k}={ }^{10} / 17$, and $\mathrm{E} X=10+\mathrm{EI}_{1}+\mathrm{EI}_{2}+\ldots$ $+\mathrm{EI}_{8}=10+8 \cdot{ }^{10} / 17=14^{12} / 17 \approx 14.7$.

## Answer

a) 0; b) $1120 / 2431$; c) $14^{12} / 17$.

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