1. What is the probability that in a random sequence of 8 ones and two zeros, there are exactly three ones between the two zeros?
Answer: $\frac{2}{15}$.

Solution: Number the positions in the sequence from the first to the tenth. If one of the zeros is in positions $1-4$ or $7-10$, then it can have only one zero that is three digits away. If one of the zeros is in position 5 or 6, then it has two such zeros. The number of all outcomes is $10 \cdot 9=90$. The number of favorable outcomes is $8+2 \cdot 2=12$. The probability is $\frac{12}{90}=\frac{2}{15}$.

Comment: Correct solution - 20 points. Correct solution with arithmetic errors - 15 points. Incorrect method of counting the numerator or denominator - 5 points. Solution started but progress is insignificant - 1 point. Only the correct answer without a solution - 0 points.