Probability of Occurrence of At Least One Independent Event

Theorem
Let $$\mathcal E = \left({\Omega, \Sigma, \Pr}\right)$$ be a probability space.

Let $$A_1, A_2, \ldots, A_m \in \Sigma$$ be independent events in the event space of $$\mathcal E$$.

Then the probability of at least one of $$A_1$$ to $$A_m$$ occurring is:
 * $$1 - \prod_{i=1}^m \left({1 - \Pr \left({A_i}\right)}\right)$$

Corollary
Let $$A$$ be an event in an event space of an experiment $$\mathcal E$$ whose probability space is $$\left({\Omega, \Sigma, \Pr}\right)$$.

Let $$\Pr \left({A}\right) = p$$.

Suppose that the nature of $$\mathcal E$$ is that its outcome is independent of previous trials of $$\mathcal E$$.

Then the probability that $$A$$ occurs at least once during the course of $$m$$ trials of $$\mathcal E$$ is $$1 - \left({1 - p}\right)^m$$.

Proof
Follows as a direct result of Probability of Independent Events Not Happening.

Let $$B$$ be the event "None of $$A_1$$ to $$A_m$$ happen".

From Probability of Independent Events Not Happening:
 * $$\Pr \left({B}\right) = \prod_{i=1}^m \left({1 - \Pr \left({A_i}\right)}\right)$$

Then $$\Omega \setminus B$$ is the event "Not none of $$A_1$$ to $$A_m$$ happen", or "At least one of $$A_1$$ to $$A_m$$ happens".

From Elementary Properties of Probability Measure:
 * $$\forall A \in \Omega: \Pr \left({\Omega \setminus A}\right) = 1 - \Pr \left({A}\right)$$

Hence the probability that at least one of $$A_1$$ to $$A_m$$ happen is:
 * $$1 - \Pr \left({B}\right) = 1 - \prod_{i=1}^m \left({1 - \Pr \left({A_i}\right)}\right)$$

Proof of Corollary
It can immediately be seen that this is an instance of the main result with all of $$A_1, A_2, \ldots, A_m$$ being instances of $$A$$.

The result follows directly.

Comment
This is a classic result which contradicts the following equally classic fallacy:


 * "There is a one in six chance of throwing a six with a single throw of a die.
 * Therefore, there is a two in six change of throwing a six on two throws of a die."

In fact this is an example of "occurrence of at least one independent event".

The probability of throwing at least one six on two throws of a die is in fact:
 * $$1 - \left({1 - \frac 1 6}\right)^2 = \frac {11} {36} < \frac 2 6$$

Not a lot in it, but definitely significantly less.

See De Méré's Paradox for a real-world application of this result as it occurred in history.