Mr. M receive $10$ messages on New Years Eve on average. Determine the probability that he receives $3$ messages at most on the next New Years Eve. 
Mr. M receive $10$ messages on New Years Eve on average. Determine the probability that he receives $3$ messages at most on the next New Years Eve.

So, "on average" should mean that $E(X) = 10$, with $X$ being a random variable that counts the number of messages Mr. M received by $n$ persons who could potentially send him such a message. It might be useful to define
$$X = X_1 \ + \ ... \ + X_n$$
with $X_k := 1$, if the $k$-th person sends Mr. M a message, $0$ otherwise. Then, each $X_k$ is Bernoulli distributed.
All in all, we are looking for the probability that
$$P(X \le 3).$$
Since each $X_k$ is Bernoulli distributed, $X$ itself is Binomial Distributed, hence,
$$P(X \le 3) = \sum_{k = 0}^3 {n \choose k} p^k q^n{n - k}$$
with $p$ being the probability that a person sends him a message. But I cannot see how to determine $p$ with the help of the information that $E(X) = 10$. 

It is reasonable to suppose that the number of people who could potentially send him a message is quite large, and the probability to do it for any person is quite small. Binomial distribution in this case can be approximated by Poisson. Remind that the Poisson distribution serves as the limiting distribution of the number of rare events in the large series of independent trials. Here is just the case.  $\lambda=np=10$ and 
$$
P(X\leq 3)=e^{-10}\left(1+10+\frac{10^2}{2!}+\frac{10^3}{3!}\right).$$ 
