Problem 2. (Option 2)

In a box, there are 3 red, 4 gold, and 5 silver stars. Randomly, one star is taken from the box and hung on the Christmas tree. What is the probability that a red star will end up on the top of the tree, there will be no more red stars on the tree, and there will be exactly 3 gold stars, if a total of 6 stars are taken from the box?
Solution: There are a total of $3+4+5=12$ toys. It is necessary for a red toy to be on top - the probability is $\frac{3}{12}$. From the remaining 11, 3 golden, 2 silver, and 0 red toys are chosen, so the probability is $\frac{C_{4}^{3} \cdot C_{5}^{2} \cdot C_{2}^{0}}{C_{11}^{5}}$.

The desired probability is obtained by multiplying the two obtained: $\frac{3 \cdot C_{4}^{3} \cdot C_{5}^{2} \cdot C_{2}^{0}}{12 \cdot C_{11}^{5}}$

We will calculate these expressions separately: $C_{2}^{0}=1, \quad C_{5}^{2}=\frac{5 \cdot 4}{1 \cdot 2}=10$, $C_{4}^{3}=4, C_{11}^{5}=\frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}=462$.

Substitute the obtained results into the answer: $\frac{3 \cdot 4 \cdot 10 \cdot 1}{12 \cdot 462}=\frac{5}{231}$.

Answer: $\frac{5}{231}$.