2. The Chairman is selected by a fair draw from the delegates of the chosen city each day.

In a private conversation with the prime minister, the president explained that the new procedure would increase the probability that both days' Chairmen would be loyal members of the ARP. The prime minister was against the changes and countered that the new procedure would increase the probability that both days' Chairmen would be members of the hated ADP.

Who is right - the president or the prime minister?
Solution. Let $p_{k}$ be the probability that a member of the APP will be chosen from the delegation of the $k$-th city. Under the first method, the probability that the Chairperson will be a member of the APP on both days is $p=\left(\frac{p_{1}+\ldots+p_{5}}{5}\right)^{2}$. Under the second method, it is $p^{\prime}=\frac{p_{1}^{2}+\ldots+p_{5}^{2}}{5}$. From the inequality between the arithmetic mean and the quadratic mean, it follows that $p \leq p^{\prime}$, and since not all $p_{k}$ are equal, then $p<p^{\prime}$.[^2]

## $\mathrm{XI}$ Internet Olympiad in Probability Theory and Statistics.

Main (correspondence) round. 15.12.77-21.01.18

Similarly, we can verify that the Prime Minister's statement is also correct. Let $q_{k}=1-p_{k}$ be the probability that a member of the ADP will be chosen from the delegation of the $k$-th city. The probabilities that the Chairperson will be a representative of the ADP on both days are $q=\left(\frac{q_{1}+\ldots+q_{5}}{5}\right)^{2}$ under the old rules and $q^{\prime}=\frac{q_{1}^{2}+\ldots+q_{5}^{2}}{5}$ under the new rules. Again, the second probability is greater: $q<q^{\prime}$. There is no contradiction: under the second method, the probability that different parties will chair on different days decreases.

Answer: both are correct.