Give the trials the numbers $1,2,\dots,1000$.
Let $X_i$ take value $1$ if the event is occurring at the $i$-th trial and value $0$ otherwise.
$$X:=X_1+\cdots+X_{1000}$$ is the number of events that occur. This with:
$$\mathbb E(X_i)=1\times P(X_i=1)+0\times P(X_i=0)=1\times\frac{1}{1000}+0\times\frac{999}{1000}=\frac{1}{1000}$$
for each $i\in\{1,\dots.1000\}$ and:$$\mathbb E(X)=\mathbb E(X_1+\cdots+X_{1000})=\mathbb E(X_1)+\cdots+\mathbb E(X_{1000})=\frac{1}{1000}+\cdots+\frac{1}{1000}=1$$
If an event happens with probability $p$, then the probability of it not happening is $1-p$. Instead of looking at the question "does the event occur at least once?", consider its complement. That is, "what is the probability of the event never occurring?"
If we repeat the event $n$ times, the probability of the event never happening is $(1-p)^n$. Thus, the probability of it happening at least once is $1-(1-p)^n$.
Applied to your scenario, we have $p=0.015$, $n=15$. Thus...
The reason that we can't just multiply the probability of one occurrence by the number of samples is best illustrated by an example: Suppose we had $p=1.5\%$, as in your problem. If we applied to $1000$ schools, what is the probability that we get in? Saying $P[\text{I get in}] = 1000\cdot 1.5\% = 1500\%$ doesn't make since--probabilities must be between $0$ and $1$!
EDIT: I wrote this up, and then saw André's comment on the main post. The assumption of independence is key, here. If the probability of the event changes every time you perform an experiment, the answer is a bit more involved.
Best Answer
Hints:
The probability it happens $k$ times in $n$ independent attempts is $P(k)={n \choose k}p^k (1-p)^{n-k}$. Your attempts seem to have ignored the binomial co-efficients which count the different possible orderings of successes.
To answer the questions, consider: