I need some clarification for why the probability of the union of three events is equal to the right side in the following:

$$P(E\cup F\cup G)=P(E)+P(F)+P(G)-P(E\cap F)-P(E\cap G)-P(F\cap G)+P(E\cap F\cap G)$$

What I don't understand is, why is the last term(intersection of all) added back just once, when it was subtracted three times as it appears from a Venn Diagram?

Here on page 3, this is explained but not in enough details that I can understand it: http://www.math.dartmouth.edu/archive/m19w03/public_html/Section6-2.pdf


3 Answers 3


One of the axioms of probability is that if $A_1, A_2, \dots$ are disjoint, then

$$\begin{align} \mathbb{P}\left(\bigcup_{i=1}^{\infty}A_i\right) = \sum\limits_{i=1}^{\infty}\mathbb{P}\left(A_i\right)\text{.}\tag{*} \end{align}$$

It so happens that this is also true if you have a finite number of disjoint events. If you're interested in more detail, consult a measure-theoretic probability textbook.

Let's motivate the proof for the probability of the union of three events by using this axiom to prove the probability of the union of two events.

Theorem. For two events $A$ and $B$, $\mathbb{P}\left(A \cup B\right) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)$.

Proof. Write $$A \cup B = \left(A \cap B\right) \cup \left(A \cap B^{c}\right) \cup \left(A^{c} \cap B\right)\text{.}$$ Notice also that $A = \left(A \cap B^{c}\right) \cup\left(A \cap B\right)$ and $B = \left(B \cap A^{c}\right) \cup \left(A \cap B\right)$.

enter image description here

Since we have written $A$ and $B$ as disjoint unions of sets, applying (*) in the finite case, we have that $$\begin{align} \mathbb{P}\left(A\right) &= \mathbb{P}\left(A \cap B^{c}\right) + \mathbb{P}\left(A \cap B\right) \\ \mathbb{P}\left(B\right) &= \mathbb{P}\left(B \cap A^{c}\right) + \mathbb{P}\left(A \cap B\right) \\ \end{align}$$ Similarly, since $A \cup B = \left(A \cap B\right) \cup \left(A \cap B^{c}\right) \cup \left(A^{c} \cap B\right)$ is a disjoint union of sets, $$\begin{align} \mathbb{P}\left(A \cup B\right) &= \mathbb{P}\left[ \left(A \cap B\right) \cup \left(A \cap B^{c}\right) \cup \left(A^{c} \cap B\right) \right] \\ &= \overbrace{\mathbb{P}\left(A \cap B\right) + \mathbb{P}\left(A \cap B^{c}\right)}^{\mathbb{P}(A)} + \mathbb{P}\left(A^{c} \cap B\right) \\ &= \mathbb{P}\left(A\right) + \overbrace{\mathbb{P}\left(A^{c} \cap B\right)}^{\mathbb{P}(B)-\mathbb{P}(A \cap B)} \\ &= \mathbb{P}\left(A\right) + \mathbb{P}\left(B\right) - \mathbb{P}\left(A \cap B\right)\text{. } \square \end{align}$$

Now, armed with the result that we proved in the previous theorem, we can now prove the result for the probability of the union of three events.

Theorem. $\mathbb{P}\left(A \cup B \cup C\right) = \mathbb{P}\left(A\right) + \mathbb{P}\left(B\right) + \mathbb{P}\left(C\right) - \mathbb{P}\left(A \cap B\right) - \mathbb{P}\left(A \cap C\right) - \mathbb{P}\left(B \cap C\right) + \mathbb{P}\left(A \cap B \cap C\right)$

Proof. Since $A \cup B \cup C = \left(A \cup B\right) \cup C$, by the previous theorem, $$\begin{align} \mathbb{P}\left(A \cup B \cup C\right) &= \mathbb{P}((A \cup B)\cup C) \\ &= \overbrace{\mathbb{P}\left(A \cup B\right) + \mathbb{P}\left(C\right) - \mathbb{P}\left[\left(A \cup B\right) \cap C\right]}^{\text{applying the previous theorem to }\mathbb{P}((A \cup B)\cup C)} \\ &= \overbrace{\mathbb{P}\left(A\right) + \mathbb{P}\left(B\right) - \mathbb{P}\left(A \cap B\right)}^{\mathbb{P}\left(A \cup B\right) \text{ from the previous theorem}} + \mathbb{P}\left(C\right) - \mathbb{P}\left[\overbrace{\left(A \cap C\right) \cup \left(B \cap C\right)}^{(A \cup B)\cap C\text{ (distributive property of sets)}}\right] \\ &= \mathbb{P}\left(A\right) + \mathbb{P}\left(B\right) - \mathbb{P}\left(A \cap B\right) + \mathbb{P}\left(C\right) \\ &\qquad- \overbrace{\Big[\mathbb{P}\left(A \cap C\right) + \mathbb{P}\left(B \cap C\right) - \mathbb{P}\left[\left(A \cap C\right) \cap \left(B \cap C\right) \right]\Big]}^{\text{applying the previous theorem to }\mathbb{P}\left(\left(A \cap C\right) \cup \left(B \cap C\right)\right)}\text{,} \end{align}$$ and since $\left(A \cap C\right) \cap \left(B \cap C\right) = A \cap B \cap C$, the result is proven. $\square$

  • $\begingroup$ While I didn't ask for an analytic clarification explicitly, I've been trying to do this during the past 2 days, so I thank you for this. But how did you turn P[(A∪B)∩C] to P[(A∩C)∪(B∩C)]? $\endgroup$
    – Hex4869
    Feb 9, 2014 at 7:18
  • 1
    $\begingroup$ That is the distributive property of sets, usually proven in a Real Analysis class. See en.wikipedia.org/wiki/… . Note that $\left(A \cup B\right) \cap C = C \cap \left(A \cup B\right)$ as you look through this link. $\endgroup$ Feb 9, 2014 at 7:21
  • $\begingroup$ I'll go with this answer because it helped me in several ways. $\endgroup$
    – Hex4869
    Feb 10, 2014 at 6:52
  • $\begingroup$ To comment on why P[ (A ∪ B) ∩ C ] is able to become P[ (A ∩ C) ∪ (B ∩ C) ], try drawing out a Venn Diagram to understand visually what's happening here. $\endgroup$ Apr 21, 2018 at 19:54
  • 2
    $\begingroup$ @Clarinetist amazing drawing you got there $\endgroup$
    – Christy
    Sep 4, 2018 at 12:26

3 sets

Since in this formula you add this part 3 times(in the first 3 terms) and subtract it also 3 times(in the next 3 terms), you have to add it again.


An interesting way to find the answer is by looking at the following Venn diagram:

Venn diagram: Union of 3 events

The numbers in red indicate how many times a certain section has been included/added by using $P(A) + P(B) + P(C)$. The sections which feature more than once must be subtracted from the former to avoid double-counting and obtain the final answer.

  1. The blank sections are added once. $\implies$ No subtraction required.

  2. The dashed (||) sections represent $P(A \cap B \cap C^c)$, $P(A \cap B^c \cap C)$, and $P(A^c \cap B \cap C)$. Each term is included twice, $\implies$ hence each must be subtracted once.

  3. The dotted (::) section represents $P(A \cap B \cap C)$. Appears thrice $\implies$ must subtracted twice.

This yields:

$$P(A \cup B \cup C ) = P(A) + P(B) + P(C) - P(A \cap B \cap C^c) - P(A \cap B^c \cap C) - P(A^c \cap B \cap C) - 2P(A \cap B \cap C).$$

Observing from the figure that: $$P(A \cap B \cap C^c) = P(A \cap B) - P(A \cap B \cap C),$$ $$P(A \cap B^c \cap C) = P(A \cap C) - P(A \cap B \cap C),$$ $$P(A^c \cap B \cap C) = P(B \cap C) -P(A \cap B \cap C),$$

Replacing these values in the above results in the final formula, given by: $$P(A \cup B \cup C ) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C).$$

  • $\begingroup$ I finally understood it, thanks! $\endgroup$
    – antimirov
    Apr 11, 2018 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.