4. Pete throws a die several times onto the table and counts the sum of the points that come up. For any natural number $n$, the event $A_{n}$ occurs if this sum equals $n$. Find the probability of the event $A_{10}$.
Answer: $P(A)=\frac{1}{9} \cdot\left(\frac{3}{6^{2}}+\frac{27}{6^{3}}+\frac{80}{6^{4}}+\frac{126}{6^{5}}+\frac{126}{6^{6}}+\frac{84}{6^{7}}+\frac{36}{6^{8}}+\frac{9}{6^{9}}+\frac{1}{6^{10}}\right)$