In a country with $N$ people, if everyone calls one random person in the country, what is the expected number of people who dont receive a call?
If $f(m,n,a,b)=$Expected fraction of people who receive $t$ calls, for $a
In a country with $N$ people, if everyone calls one random person in the country, what is the expected number of people who dont receive a call?
If $f(m,n,a,b)=$Expected fraction of people who receive $t$ calls, for $a
This, and Francesco's answer, are based on the question as originally posed. Set $X=1$ or $X=m$ and $N=n$ to fit the current statement.
Assuming that no one calls themself, nor calls anyone else more than once, the probability of receiving exactly $k$ calls is $ \binom{N-1}{k}\left(\frac{X}{N-1}\right)^k\left(1-\frac{X}{N-1}\right)^{N-k-1}\tag{1} $ Thus, the probability of receiving no calls is $ \left(1-\frac{X}{N-1}\right)^{N-1}\tag{2} $ By linearity of expectation, the expected number of people who don't get a call is $ N\left(1-\frac{X}{N-1}\right)^{N-1}\tag{3} $ As $N\to\infty$, $(1)$ becomes $ \frac{X^k}{k!}e^{-X}\tag{4} $ which says that the probability of getting $k$ calls follows a Poisson distribution with mean $X$.
By linearity of expectation, the expected number of people who get exactly $k$ calls is $N$ times $(1)$: $ N\binom{N-1}{k}\left(\frac{X}{N-1}\right)^k\left(1-\frac{X}{N-1}\right)^{N-k-1}\tag{5} $ Summing over $a
The probability of not receiving a call is $(1 - \frac{X}{N-1})^{N-1}$ (assuming uniform distribution of the calls, that everybody has a phone number and that one doesn't call himself). The expected number of people which will not receive a call is thus obtained by summing over all the N people in the country.
Do you recognize a well known limit?
ps: This seems like a homework question so I tried to give only a hint.