Problem: A random number selector can only select one of the nine integers $1,2, \ldots, 9$, and it makes these selections with equal probability. Determine the probability that after $n$ selections ( $n>1$ ), the product of the $n$ numbers selected will be divisible by 10. Solution: This is simply the probability that there exists $2,4,6,8$ in the first $n$ selections and there exists $5$ in the first $n$ selections. If we let the former event be $A$ and the ladder be $B$, we know that $ \begin{aligned} P(A \wedge B) & =1-P(\neg A \vee \neg B) \\ & =1-[P(\neg A)+P(\neg B)-P(\neg A \wedge \neg B)] \\ & =1-\left[(5 / 9)^n+(8 / 9)^n-P(\text {no even AND no } 5)\right] .\end{aligned} $We can also compute that $P(\text {no even AND no } 5) = (\frac{4}{9})^n$. Hence, the answer is $1 - \left( \frac{5}{9} \right)^{n} + \left( \frac{8}{9} \right)^{n} - (\frac{4}{9})^{n}.$