Average number of events happening if each happens with $p=\frac{1}{n}$ and we run it $10000 n$ times. Let an event $e$ have probability of happening $\frac{1}{n}$.
Let us assume we have $m$ independent possibilities for similar events to happen. With $m>>n$. What is the average number of times the event will happen?
The question came out of a discussion on facebook respect to the meteorite and someone claiming that the probability of being hit by an asteroid being $\frac{1}{700000}$. The probability was seen as too high considering there are $7$ billion people in the world.
So in this case let $n=700000; m=7000000000$. What is the average number of time that an event should happen if each time it has $\frac{1}{n}$ possibility of happening and we extract this probability $10000n$.
 A: It should happen $m\over n$ times. In your case, $10000$ times.
A: Note that in linguistics "chance of being hit by an asteroid" is commonly confused with "chance an asteroid hits at least one person". I believe that's when the confusion arises (as other users pointed out, if we interpret the comment as you did every asteroid should hit $10^4$ people on average, and since that definitely doesn't happen the $\frac{1}{700000}$ wouldn't make any sense).
Now, if you toss a coin whose result "head" has probability $p$, and you do it $m$ times, you get what is called the binomial distribution, which states that (naming H = number of heads on $m$ trials)
$$P(H=k) = \binom{m}{k}p^k (1-p)^{m-k}$$
In your example you want to calculate the chance of a specific asteroid hitting someone (=at least one person)... in your example $p=\frac{1}{700000}$ and $m=1$ ! So the chance is indeed really small ($p$).
A: Suppose there are $m = 7\times 10^9$ people in the world,
each of whom has the same chance to be struck by a meteor,
and for each individual person the the probability that that person
will be struck by a meteor is $1/n$ where $n = 7\times 10^5$.
Assign a number from $1$ to $m$ to each person, and for each integer $k$
in that interval ($1 \leq k \leq m$)
let $X_k$ be random variable defined so that 
$X_k = 1$ if person number $k$ is hit by a meteor, $X_k = 0$ otherwise.
Then by simple calculation, the expected value of $X_k$ is
$$
E(X_k) = 0 \cdot P(X_k = 0) +  1 \cdot P(X_k = 1) = P(X_k = 1) = \frac1n.
$$
The expected value is the probabilistic "average" of a random variable,
and it has the very nice property that for any two random variables
$X$ and $Y$, if expected value is defined for each of those variables
then $E(X + Y) = E(X) + E(Y)$.
This is true even if $X$ and $Y$ are not independent events.
It is also true for sums of more than two variables, as long as only a finite
number of variables is involved.
To find out how many people in the world are struck by a meteor,
we simply go through the entire list of people and add $1$ to
"number of people struck" for each person who is struck, $0$ for anyone else.
That is, we add up the sum
$X_1 + X_2 + X_3 + \ldots + X_m$.
But by the additive property of expected value,
\begin{align}
E(X_1 + X_2 + X_3 + \ldots + X_m) 
&= E(X_1) + E(X_2) + E(X_3) + \ldots + E(X_m)\\
&= \overbrace{\frac1n + \frac1n + \ldots + \frac1n}^{\text{$m$ terms}} \\
&= \frac mn
\end{align}
and that is the expected (i.e. average) number of people who
will be struck by meteors.
