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:

$\displaystyle 1 - \prod_{i=1}^m \left({1 - \Pr \left({A_i}\right)}\right)$

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:

$\displaystyle \Pr \left({B}\right) = \prod_{i=1}^m \left({1 - \Pr \left({A_i}\right)}\right)$

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:

$\displaystyle 1 - \Pr \left({B}\right) = 1 - \prod_{i=1}^m \left({1 - \Pr \left({A_i}\right)}\right)$

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.

$\blacksquare$

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:

$1 - \left({1 - \dfrac 1 6}\right)^2 = \dfrac {11} {36} < \dfrac 2 6$

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.