B2. In a bag, there are 5 red and 5 green coins. Each coin has the number 0 on one side and the number 1 on the other side. All coins are fair, meaning the probability of landing on 0 is equal to the probability of landing on 1. We randomly select 6 coins from the bag and toss them. What is the probability that the sum of the numbers that land on the red coins is equal to the sum of the numbers that land on the green coins?
B2. If the coins are fair, the probability can be calculated using the formula

$$
P=\frac{\text { favorable outcomes }}{\text { all outcomes }}
$$

First, let's count all the outcomes. We get them by first choosing 6 coins from the urn in $\binom{10}{6}$ ways and then flipping each coin in 2 ways. Therefore, the total number of outcomes is $\binom{10}{6} \cdot 2^{6}=13440$.

To count the favorable outcomes, we consider the cases of how many red coins are drawn and how many of these coins land on 1.

If we draw 5 red coins, this can be done in $\binom{5}{5}=1$ way, and the remaining 1 green coin can be chosen in $\binom{5}{1}=5$ ways. Since there is only one green coin, the number 1 can land on the red coins either 0 times or 1 time. The first possibility can occur in $\binom{5}{0}=1$ way, and the second in $\binom{5}{1}=5$ ways. In both cases, the green coin is uniquely determined. This gives us a total of

$$
1 \cdot 5 \cdot(1+5)=30
$$

outcomes.

If we draw 4 red coins, this can be done in $\binom{5}{4}=5$ ways, and the remaining 2 green coins can be chosen in $\binom{5}{2}=10$ ways. Then, the number 1 can land on the red coins 0 times, 1 time, or 2 times. The first possibility can occur in $\binom{4}{0}=1$ way, as the green coins are then uniquely determined. The second possibility can occur in $\binom{4}{1} \cdot\binom{2}{1}=8$ ways, as the number 1 must land on one of the four red coins and one of the two green coins. The last possibility can occur in $\binom{4}{2}=6$ ways, as the green coins are then uniquely determined. This gives us a total of

$$
5 \cdot 10 \cdot(1+8+6)=750
$$

outcomes.

If we draw 3 red coins, this can be done in $\binom{5}{3}=10$ ways, and the remaining 3 green coins can be chosen in $\binom{5}{3}=10$ ways. The number 1 can land on the 3 red coins 0 times, 1 time, 2 times, or 3 times, and the same number of times on the 3 green coins. The first possibility can occur in $\binom{3}{0}^{2}=1$ way, the second in $\binom{3}{1}^{2}=9$ ways, the third in $\binom{3}{2}^{2}=9$ ways, and the last in $\binom{3}{3}^{2}=1$ way. Therefore, we have a total of

$$
10 \cdot 10 \cdot(1+9+9+1)=2000
$$

outcomes.

Due to the symmetry between red and green coins, the number of favorable outcomes when drawing 2 or 1 red coin is the same as when drawing 2 or 1 green coin, which is the same as drawing 4 or 5 red coins. Therefore, the total number of favorable outcomes is

$$
30+750+2000+750+30=3560
$$

The probability that the sum of the numbers that land on the red coins is equal to the sum of the numbers that land on the green coins is therefore

$$
P=\frac{3560}{13440}=\frac{89}{336} .
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_e347356273e0c1397da7g-30.jpg?height=75&width=1751&top_left_y=1136&top_left_x=164)

![](https://cdn.mathpix.com/cropped/2024_06_07_e347356273e0c1397da7g-30.jpg?height=66&width=1731&top_left_y=1192&top_left_x=174)

Listing all possible color combinations of the drawn coins ...................................... 1 point.

![](https://cdn.mathpix.com/cropped/2024_06_07_e347356273e0c1397da7g-30.jpg?height=54&width=1727&top_left_y=1304&top_left_x=176)

Correct final result .............................................................................................. 1 point.