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:$$\mathbb E(X)=\mathbb E(X_1+\cdots+X_{1000})=\mathbb E(X_1)+\cdots+\mathbb E(X_{1000})=\frac{1}{1000}+\cdots+\frac{1}{1000}=1$$
- 2 events $A$ and $B$ are independent if
$$P(A)P(B) = P(A \cap B)$$
- 3 events $A,B,C$ are independent if
$$P(A)P(B) = P(A \cap B)$$
$$P(A)P(C) = P(A \cap C)$$
$$P(C)P(B) = P(C \cap B)$$
$$P(A)P(B)P(C) = P(A \cap B \cap C)$$
If we have the 1st 3 but not the 4th, then $A,B,C$ are not independent, but they are pairwise independent.
- So what's the probability of $A \cup B \cup C$? The probability at least one occurs is equal to 1 - the probability that none occur.
$$P(A \cup B \cup C) = 1-P(A^c \cap B^c \cap C^c)$$
$$ = 1-P(A^c)P(B^c)P(C^c) \tag{***}$$
$$ = 1-(1-P(A))(1-P(B))(1-P(C))$$
$$ = 1-(1-p)(1-p)(1-p)$$
$$ = 1-(1-p)^3$$
What just happened at $(***)$?
Actually, if $A$ and $B$ are independent, then
- $A^C$ and $B^C$ are independent
- $A^C$ and $B$ are independent
- $A$ and $B^C$ are independent
Similarly, if $A$, $B$ and $C$ are independent, then $A^C, B^C, C^C$ are independent.
Hence
$$P(A^C)P(B^C) = P(A^C \cap B^C)$$
$$P(A^C)P(C^C) = P(A^C \cap C^C)$$
$$P(C^C)P(B^C) = P(C^C \cap B^C)$$
$$P(A^C)P(B^C)P(C^C) = P(A^C \cap B^C \cap C^C)$$
The last part is what is used to justify $(***)$.
Now finally how about for $n$ independent events $A_1, A_2, ..., A_n$?
$$P(A_1 \cup ... \cup A_n) = 1-P(A_1^c \cap ... \cap A_n^c)$$
$$ = 1-P(A_1^c)...P(A_n^c)$$
$$ = 1-(1-P(A_1))...(1-P(A_n))$$
$$ = 1-(1-p)...(1-p)$$
$$ = 1-(1-p)^n$$
It looks like you meant to compute probability exactly 1 occurs, probability exactly 2 events occur, etc.
To compute probability exactly 1 occurs:
$$P(A_1 \cap A_2^C \cap ... \cap A_n^C) = P(A_1)P(A_2^C) ... P(A_n^C) = p(1-p)^{n-1}$$
Similarly, we have
$$P(A_1^C \cap A_2 \cap ... \cap A_n^C) = p(1-p)^{n-1}$$
etc
Thus we have
Probability exactly 1 occurs = $np(1-p)^{n-1}$, not $p$
Note that
$$np(1-p)^{n-1} = \binom{n}{1}p^1(1-p)^{n-1}$$
Similarly, probability exactly 2 occur = $\binom{n}{2}p^2(1-p)^{n-2}$
Thus probability any $n$ occur =
$$\binom{n}{1}p^1(1-p)^{n-1} + \binom{n}{2}p^2(1-p)^{n-2} + ... + \binom{n}{n}p^n(1-p)^{n-n}$$
By binomial expansion, we have:
$$= ((p) + (1-p))^n - \binom{n}{0}p^0(1-p)^{n-0} = 1 - (1-p)^n$$
Best Answer
For any future visitors, the expression given for the probability of at least $m$ out of $n$ events $(A_1, \ldots ,A_n)$ in An Introduction to Probability by William Feller is this:
$$P_m = S_m - \binom{m}{1}S_{m+1}+\binom{m+1}{2}S_{m+2}-\ldots\pm\binom{n-1}{m-1}S_n$$
where $$\displaystyle S_k = \sum_{1\leq i_1< i_2\ldots< i_k\leq n}P(A_{i_1}\cap A_{i_2}\cap\ldots \cap A_{i_k})$$
It's certainly not pretty, but it is general.
It is proved by finding the expression for exactly $m$ events and then adding the expressions from $m$ to all $n$ using induction.