[ Inclusion-Exclusion Principle ]  [ Recurrence relations (other).]

An implausible legend says that one day Stirling was contemplating Stirling numbers of the second kind and, in a pensive mood, was tossing 10 fair dice onto the table. After another throw, he suddenly noticed that the combination of points that came up included all numbers from 1 to 6. At that moment, Stirling began to wonder what the probability of such an event would be. What is the probability that when 10 dice are rolled, each number of points from 1 to 6 appears at least once?
Let's find the probability of event $A$ "some number is missing in the rolled combination". Denote $A_{k}$ as the event "number $k$ is missing". Then

$A=A_{1} \cup A_{2} \cup A_{3} \cup A_{4} \cup A_{5} \cup A_{6}$.

Using the inclusion-exclusion principle, we get:

$\mathrm{P}(A)=\mathrm{P}\left(A_{1}\right)+\mathrm{P}\left(A_{2}\right)+\ldots+\mathrm{P}\left(A_{6}\right)-\mathrm{P}\left(A_{1} \cap A_{2}\right)-\mathrm{P}\left(A_{1} \cap A_{3}\right)-\ldots-\mathrm{P}\left(A_{5} \cap A_{6}\right)+\mathrm{P}\left(A_{1} \cap A_{2} \cap A_{3}\right)+\mathrm{P}\left(A_{1} \cap A_{2} \cap A_{4}\right)+\ldots+$

$\mathrm{P}\left(A_{4} \cap A_{5} \cap A_{6}\right)-\ldots-\mathrm{P}\left(A_{1} \cap A_{2} \cap A_{3} \cap A_{4} \cap A_{5} \cap A_{6}\right)$.

All probabilities $\mathrm{P}\left(A_{k}\right)$ are equal to $(5 / 6)^{10}$, and there are $C_{6}^{1}=6$ of them. The probabilities of pairwise intersections are $\left({ }^{4} / 6\right)^{10}$, and there are $C_{6}^{2}=15$ of them, and so on. Therefore,

$P(A)=6 \cdot(5 / 6)^{10}-15 \cdot\left(\frac{4}{6}\right)^{10}+20 \cdot\left(\frac{3}{6}\right)^{10}-15 \cdot(2 / 6)^{10}+6 \cdot\left(\frac{1}{6}\right)^{10}$.

Thus, the probability that each number appears at least once is $1-\mathrm{P}(A)$.

## Answer

$1-6 \cdot(5 / 6)^{10}+15 \cdot(4 / 6)^{10}-20 \cdot(3 / 6)^{10}+15 \cdot(2 / 6)^{10}-6 \cdot\left(\frac{1}{6}\right)^{10} \approx 0.272$.