Probability that any $n$ independent events occur Say there are n independent events, each with a probability p of occurring. What is the probability that any of these events occur?
My guess is that the probability is
$$ P(any) = p(1) + p(2) + p(3)+... + p(n)$$
$$ P(any) = p + p^2 + p^3 +... + p^n$$
$$ P(any) = \sum_{i=1}^n p^i$$
Does this sum have another form that can be computed easily? For example, if $n=800$, I don't want to actually calculate all $800$ terms.
 A: *

*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:
$$= ((p) + (1-p))^n - \binom{n}{0}p^0(1-p)^{n-0} = 1 - (1-p)^n$$
A: The best way to do this is to use $1-\prod_k(1-p_k)$. The part after the $1-$ finds the probability that none of the events happen, and the $1-$ part finds the opposite of that, i.e. the probability that at least one of these events happen.
A: Note that $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
So if $A$ and $B$ are independent, then $P(A\cap B)=P(A)P(B)$.
This is because more than one of the events can also happen together.
For three events:
$$P (A \cup B \cup C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)$$
and similarly for more events.
So the correct way to deal with this situation is to calculate the complimentary probability $$P(\overline{A\cup B \cdots})=P(\bar{A}\cap \bar{B} \cap \cdots )$$ and then your desired probability is
$$P(\text{any event})=1-P(\bar{A}\cap \bar{B} \cap \cdots )$$ 
So by independence,
$$P(\text{any event})=1-P(\bar A)P(\bar B)\cdots=1-\big[(1-P(A))(1-P( B))\cdots\big]$$ 
A: Hint:
The probability that any occur is equal to one minus the probability that none occur.
The probability that no events occur is equal to the probability that the first doesn't occur AND the second doesn't occur AND ... AND the $n^{th}$ doesn't occur.
Recall the "and rule" of probability; the probability of $A$ and $B$ is equal to the probability of $A$ multiplied by the probability of $B$. (this is for independent events)
A: As others have pointed out, it is easier to just compute the complementary event: none of the n events occur. The probability of none happening is $(1-p)^n$, so the probability that any of $n$ independent events occur is  $1-(1-p)^n$
But I started this answer because I wanted to expand on the (more difficult) way you chose to compute this probability. Your method of computing the probability for exactly one event happening, exactly 2 events happening and so on, and adding them all together is correct. However the way you compute these probabilities is not correct.
Let's say you are having $n$ events (all independent and equiprobable) and you want to compute the probability that exactly one is happening. To approach the problem you can think about one specific event happening, say the first one. What's the probability of this? It is $p\times (1-p)^{n-1}$. Or in other words, the probability of the first one happening times the probabilities of the rest not happening. If we are looking for any one event (exactly one event) then we have to consider all other single events happening (the 2nd, 3rd,..nth).
These have the same probability $p\times (1-p)^{n-1}$ and we have $n$ of them.
How about exactly two events happening? If we take two specific events, then the probability is $p^2\times (1-p)^{n-2}$. And how many of these cases we have? It's the number of combinations we can have when we select $2$ items out of $n$. This is written as $n \choose 2$.
More generally the probability of exactly $i$ events happening is:
$$\binom{n}{i}\times p^i\times (1-p)^{n-i}$$
So your sum (to calculate that any event is happening) is:
$$\sum_{i=1}^n \binom{n}{i}\times p^i\times (1-p)^{n-i}$$
If this sum started from index $0$ then it would be equal to $1$ (proof using the binomial expansion formula). So our sum is $1 - (\text{term indexed 0}) = 1-(1-p)^n$
