# Probability of Occurrence of At Least One Independent Event

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## Theorem

Let $\EE = \struct {\Omega, \Sigma, \Pr}$ be a probability space.

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

Then the probability of at least one of $A_1$ to $A_m$ occurring is:

- $\ds 1 - \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$

### Corollary

Let $A$ be an event in an event space of an experiment $\EE$ whose probability space is $\struct {\Omega, \Sigma, \Pr}$.

Let $\map \Pr A = p$.

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

Then the probability that $A$ occurs at least once during the course of $m$ trials of $\EE$ is $1 - \paren {1 - p}^m$.

## 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:

- $\ds \map \Pr B = \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$

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:

- $\forall A \in \Omega: \map \Pr {\Omega \setminus A} = 1 - \map \Pr A$

Hence the probability that at least one of $A_1$ to $A_m$ happen is:

- $\ds 1 - \map \Pr B = 1 - \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$

### 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.

$\blacksquare$

## Example

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 chance 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:

- $1 - \paren {1 - \dfrac 1 6}^2 = \dfrac {11} {36} < \dfrac 2 6$

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.