5. Multicolored Balls (from 6th grade, 1 point). In the red box, there are 100 red balls, and in the green box, there are 100 green balls. Eight red balls were moved to the green box, and then the same number of balls were moved from the green box to the red box. The balls in the boxes were thoroughly mixed. What is now greater: the probability of randomly drawing a green ball from the red box or a red ball from the green box?
Solution. After all the balls have been moved, there are still 100 balls in each box. This means that the number of red balls in the green box is the same as the number of green balls in the red box.

## Answer: the probabilities are equal.

$\mathbf{6}^{1}$. Tournament. In a tournament, 100 wrestlers of different strengths participate. In any match between two wrestlers, the stronger one always wins. In the first round, the wrestlers are randomly paired and matches are held. For the second round, the wrestlers are again randomly paired (some pairs may repeat). The prize goes to the wrestler who wins both matches. Find:

a) (from 6th grade, 1 point) the smallest possible number of tournament winners;

b) (from 8th grade, 2 points) the expected number of tournament winners.

![](https://cdn.mathpix.com/cropped/2024_05_06_e03b48fad60e6548aec7g-03.jpg?height=523&width=580&top_left_y=1986&top_left_x=1315)[^0]

Solution. Let's number the wrestlers from the weakest #1 to the strongest #100.

a) We will show that it is possible for only the strongest wrestler to be the winner. If in the first round the wrestlers are paired as follows:

$$
(1 ; 2),(3 ; 4),(5 ; 6), \ldots,(99 ; 100)
$$

then all the odd-numbered wrestlers lose in the first round. If in the second round the wrestlers are paired as follows:

$$
(2 ; 3),(4 ; 5),(6 ; 7), \ldots,(98 ; 99),(1 ; 100)
$$

then all the even-numbered wrestlers lose, except for wrestler #100. Thus, he will be the only winner of the entire tournament.

b) Let $I_{k}$ be the indicator of the event "the $k$-th wrestler won both matches." The probability of winning a match for the $k$-th wrestler is the probability that his opponent is one of the $k-1$ weaker wrestlers. Therefore, the probability of winning is $\frac{k-1}{99}$. The probability of winning both times is $\left(\frac{k-1}{99}\right)^{2}$, so

$$
\mathrm{E} I_{k}=\left(\frac{k-1}{99}\right)^{2}
$$

The total number of winners $X$ is the sum of all indicators for $1 \leq k \leq 100$, so

$$
\mathrm{E} X=\mathrm{E} I_{1}+\mathrm{E} I_{2}+\ldots+\mathrm{E} I_{100}=\left(\frac{0}{99}\right)^{2}+\left(\frac{1}{99}\right)^{2}+\ldots+\left(\frac{99}{99}\right)^{2}=\frac{0^{2}+1^{2}+2^{2}+\ldots+99^{2}}{99^{2}}
$$

The sum in the numerator can be simplified using the formula of Archimedes:

$$
1^{2}+2^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
$$

We get: $\mathrm{E} X=\frac{99 \cdot 100 \cdot 199}{6 \cdot 99^{2}}=\frac{100 \cdot 199}{6 \cdot 99} \approx 33.5$.

Answer: a) 1 ; b) 33.5 .