What is the probability that in the 5-number lottery draw, at least one of the numbers drawn was among the winning numbers from the previous week?
Solution. In the 5/90 lottery draw, five numbers are drawn from 90, which can happen in $\binom{90}{5}$ ways.

Instead of calculating the favorable cases - that is, the draws in which at least one number that appeared among the winning numbers of the previous week occurs - let's calculate the number of unfavorable cases. This happens when 5 numbers are drawn from the 85 numbers that did not appear among the winning numbers of the previous week; this can happen in $\binom{85}{5}$ ways. If we subtract this from the total number of cases, we get the number of favorable cases. The desired probability $\mathbf{P}$ is given by the ratio of the favorable cases to the total number of cases, that is,

$$
\mathbf{P}=\frac{\binom{90}{5}-\binom{85}{5}}{\binom{90}{5}}=1-\frac{\binom{85}{5}}{\binom{90}{5}}=1-\frac{\frac{85!}{5!!0!}}{\frac{90!}{5!\cdot 85!}}=1-\frac{85!\cdot 85!}{80!\cdot 90!}
$$

Simplify the fraction by $85!$ and $80!$:

$$
\mathbf{P}=1-\frac{81 \cdot 82 \cdot 83 \cdot 84 \cdot 85}{86 \cdot 87 \cdot 88 \cdot 89 \cdot 90}
$$

Factor both the numerator and the denominator into prime factors, then perform the simplifications:

$$
\mathbf{P}=1-\frac{2^{3} \cdot 3^{5} \cdot 5 \cdot 7 \cdot 17 \cdot 41 \cdot 83}{2^{5} \cdot 3^{3} \cdot 5 \cdot 11 \cdot 29 \cdot 43 \cdot 89}=1-\frac{3644613}{4883252}=\frac{1238639}{4883252} \approx 0.25365
$$

Therefore, the probability that the event will occur is approximately $0.25\%$.