What is the probability that I have seen every time on the clock? Assuming a digital clock shows only hours and minutes, there are 1440 different times it may show. If you checked the time on 35000 independent occasions, what is the probability that you would have seen all 1440 times?
 A: I'm assuming that the time of the day when we view the clock is chosen independently.
This problem is actually a bit more involved than one might expect.  Let's call our set of total times $T$; we have $T = \{1,2,\ldots ,1440\}$.  For any given $S \subset T$, define the event $A_S$ to be the event that we missed at least the times in $S$, and $B_S$ to be the event that we missed precisely the times in $S$; that is $B_S$ is the event that we missed the times in $S$ and no others.  Then we are looking for $1 - \mathbb{P}(B_\emptyset)$.  By the principle of inclusion exclusion, we have \begin{equation}
\mathbb{P}(B_\emptyset) = \sum\limits_{S \subseteq T} (-1)^{|S|} \mathbb{P}(A_{S}).  \tag{1} \end{equation}
If $S$ has $k$, elements, then the probability we never see any elements in $S$ is precisely $\left( \frac{1440 - k}{1440}\right)^{35000}.$  Since this only depends on $|S|$ and not on $S$ itself, we can convert the sum in equation $(1)$ from a sum over subsets to a sum over $k$.  For any given $k$, we have a total of $\binom{1440}{k}$ choices for $S$.  We thus have \begin{equation}
\mathbb{P}(B_\emptyset) = \sum\limits_{k=0}^{1440} (-1)^{k} \left( \frac{1440 - k}{1440}\right)^{35000} \binom{1440}{k}. \end{equation}
This means that $$ 1 - \mathbb{P}(B_\emptyset) =  \sum\limits_{k=1}^{1440} (-1)^{k+1} \left( \frac{1440 - k}{1440}\right)^{35000} \binom{1440}{k}. $$
EDIT:  According to my quick MATLAB script, summing up the first $60$ terms gives a value of $3.971247286302689 \times 10^{-8}$; the same happens when I sum the first $61$ terms, so I assume I'm hitting numerical precision.  With inclusion-exclusion, the partial sums alternately over- and under-estimate the final sum.  Thus, we know that the probability is somewhere around $3.97 \times 10^{-8}$.
