# Probability of a finite union of non-disjoint events derivation

Using only the axioms of probability, derive the following result using induction:

\begin{align} \Pr\left(\bigcup_{i=1}^\infty A_i\right) = &\sum_{i=1}^\infty P(A_i) - \sum_{i

I have began the proof by showing that it holds for $$n = 2$$, and have assumed it holds for $$n \le k$$ for some arbitrary $$k$$. Now it just comes to showing that it holds for $$k + 1$$:

\begin{align} \Pr\left(\bigcup_{i=1}^{k+1} A_i\right) &= \Pr\left(A_{k+1} \cup \bigcup_{i=1}^kA_i\right) \\ &= \Pr(A_{k+1}) + \Pr\left(\bigcup_{i=1}^k A_i\right) - \Pr\left(A_{k+1} \cap \bigcup_{i=1}^kA_i\right) \end{align}

where the last term can be written as:

$$\Pr\left(\bigcup_{i=1}^k(A_i \cap A_{k+1})\right)$$

This is where I am stuck and cannot simplify the expression to get the desired result. I am positive I am on the right track with the "proof", but do not know what result or axiom to use to continue

Any hints to how I can continue will be appreciated.

• Remember that you want to prove it for $k+1$ assuming the $k$ case is true. Assuming the $k$ case is true, you can expand out $P(\bigcup_{i=1}^k X_i)$ for any collection of $k$ sets $X_1, X_2, \ldots, X_k$. The second and third terms on the RHS of your second equation are unions of $k$ sets, so you can now expand them. – Paul Sep 27 '18 at 21:26
• Please add the [self-study] tag & read its wiki. – gung - Reinstate Monica Oct 24 '18 at 14:53
• Did @Sebastian's answer help you? If so, please consider upvoting or accepting the answer. This is done by clicking the upwards normal distribution & possibly clicking on the check mark below the vote total. If it isn't quite what you needed, consider leaving a comment below his answer. – gung - Reinstate Monica Oct 24 '18 at 14:55

## 1 Answer

You can use that you already proved the result for $$n .

Therefore $$P(\cup_{i=1}^k(A_{k+1}\cap A_i))=\sum_{i=1}^k P(A_{k+1}\cap A_i)-\sum_{i