Probability Proof $ \mathbb{P}\left(A_{1} \cup A_{2} \cup A_{3}\right) = ... $

Question

How would I go about proving this statement: $\mathbb{P}\left(A_{1} \cup A_{2} \cup A_{3}\right)=\left(\sum_{i=1}^{3} \mathbb{P}\left(A_{i}\right)\right)-\mathbb{P}\left(A_{1} \cap A_{2}\right)-\mathbb{P}\left(A_{1} \cap A_{3}\right)-\mathbb{P}\left(A_{2} \cap A_{3}\right)+\mathbb{P}\left(A_{1} \cap A_{2} \cap A_{3}\right)$

The probability space is $(\Omega,\mathbb{P})$ and for the upper equation applies $A_{i} \subseteq \Omega, i=1,2,3$

My idea is to reshape the LHS:

So we have

(i) $\left(\sum_{i=1}^{3} \mathbb{P}\left(A_{i}\right)\right)-\mathbb{P}\left(A_{1} \cap A_{2}\right)-\mathbb{P}\left(A_{1} \cap A_{3}\right)-\mathbb{P}\left(A_{2} \cap A_{3}\right)+\mathbb{P}\left(A_{1} \cap A_{2} \cap A_{3}\right)$

with

(ii) $\left(\sum_{i=1}^{3} \mathbb{P}\left(A_{i}\right)\right) = \mathbb{P}(A_{1}) \cup \mathbb{P}(A_{2}) \cup \mathbb{P}(A_{3}) $

(iii) $ \mathbb{P}(A_{1}) = \mathbb{P}\left(A_{1} \setminus A_{2}\right) + \mathbb{P}\left(A_{1} \cap A_{2}\right) <=> -\mathbb{P}\left(A_{1} \cap A_{2}\right) = \mathbb{P}\left(A_{1} \setminus A_{2}\right) - \mathbb{P}(A_{1})$

same for $\mathbb{P}(A_{2})$ and $\mathbb{P}(A_{3})$

Now we combine (i), (ii) and we get:

(iV) $\mathbb{P}(A_{1}) \cup \mathbb{P}(A_{2}) \cup \mathbb{P}(A_{3})-\mathbb{P}\left(A_{1} \cap A_{2}\right)-\mathbb{P}\left(A_{1} \cap A_{3}\right)-\mathbb{P}\left(A_{2} \cap A_{3}\right)+\mathbb{P}\left(A_{1} \cap A_{2} \cap A_{3}\right)$

Now we combine (iV), (iii) and we get:

(V) $\mathbb{P}(A_{1}) \cup \mathbb{P}(A_{2}) \cup \mathbb{P}(A_{3})+ \mathbb{P}\left(A_{1} \setminus A_{2}\right) - \mathbb{P}(A_{1})+ \mathbb{P}\left(A_{3} \setminus A_{1}\right) - \mathbb{P}(A_{3})+ \mathbb{P}\left(A_{2} \setminus A_{3}\right) - \mathbb{P}(A_{2})+\mathbb{P}\left(A_{1} \cap A_{2} \cap A_{3}\right) = \mathbb{P}\left(A_{1} \setminus A_{2}\right) + \mathbb{P}\left(A_{3} \setminus A_{1}\right) + \mathbb{P}\left(A_{2} \setminus A_{3}\right) +\mathbb{P}\left(A_{1} \cap A_{2} \cap A_{3}\right)$

Now we use this equation:

(Vi) $\mathbb{P}(A \cap B)=\mathbb{P}(A \backslash B)+\mathbb{P}(B \backslash A)+\mathbb{P}(A \cap B)$

And finally we use (VI) on (V)

$\mathbb{P}\left(A_{1} \setminus A_{2}\right) + \mathbb{P}\left(A_{3} \setminus A_{1}\right) + \mathbb{P}\left(A_{2} \setminus A_{3}\right) +\mathbb{P}\left(A_{1} \cap A_{2} \cap A_{3}\right) = \mathbb{P}\left(A_{1} \cup A_{2} \cup A_{3}\right)$

Could this work? If yes i will get drunk today :)

Answer 1

As pointed out in the comments, this identity is the probabilistic version of the Inclusion-Exclusion Principle in the special case of three sets.

Your calculation looks good to me. I'd consider proof by diagram rigorous in this setting. For example, one can read off your equation (V), namely, $$\Bbb P(A_1 \cup A_2 \cup A_3) = \color{#ff0000}{\Bbb P(A_2 \setminus A_3)} + \color{#009f00}{\Bbb P(A_3 \setminus A_1)} + \color{#3f7fff}{\Bbb P(A_1 \setminus A_2)} + \color{#7f00ff}{\Bbb P(A_1 \cap A_2 \cap A_3)} ,$$ directly from the below Venn diagram, appropriately labeled.