Consider a group of five students, each of whom chooses a day of the week (including weekends) uniformly at random to meet with their professor. The expected number of days where the professor must meet with at least one student can be expressed in the form m/n with m, n coprime. Compute m + n.
Expected value like this should involve the distribution of various cases. But I have difficulty finding the angle for cases: day, student, or professor?
expected value of number of days where prof must meet at least one student is
\begin{align*} &\sum_{d \in day} Pr(\text{professor having to meet at least on student on $d$}) \cdot 1 ~\text{day} \\ &= \sum_{d \in day}[1 - Pr(\text{professor having to meet $\textbf{no one}$ on $d$})] \cdot 1 \text{day} \qquad (2)\\ \end{align*}
The probability of the professor having to meet no one is that every student chooses a day that is not $d$. This is \begin{align*} & Pr(\text{professor having to meet $\textbf{no one}$ on $d$})] \equiv Pr(\text{student not picking day $d$)}^\text{number of students} \\ &=[1 - Pr(\text{student picking day $d$)}]^\text{number of students} \\ &= [1-1/7]^5 = (6/7)^5 \qquad (2) \end{align*}
Plugging $(2)$ back into $(1)$ gives us:
$$ \sum_{d \in day}[1 - (6/7)^5] = 7 \cdot (7^5 - 6^5)/7^5 = (7^5 - 6^5)/7^4 $$
from which one can deduce the value of $m/n$ and $m+n$