# How to show that $P(A\cup B) = P(A) + P(B) - P(A\cap B)$

Say we have a venn diagram with A on the left, intersection C in the middle and B on the right.

I want to show that $P(A\cup B) = P(A) + P(B) - P(A\cap B)$

I know what it can be written as A+B+C, but a lot of the proofs go from:

$P(A\cup B) = P(A) + P(B) + P(C)$ to

$P(A) + P(B) = P(A\cup B) + P(C)$ and then rearrange to get the proof and get $P(A\cup B) =P(A) + P(B)- P(C)$ . I just don't get how the left side of the inequality get go from $P(A\cup B)$ to $P(A) + P(B)$. Doesn't that only happen if we assume that they are disjoint. In this case, do we assume that they are disjoint? I know how to write the proof but just having a hard time actually understanding it.

• Do you mean $P(A\cup B)=P(A)+P(B)-P(C)$? – GoodDeeds Sep 15 '18 at 15:12
• Well, when I would we arrange I would get that. I fixed it so it makes more sense – George Harrison Sep 15 '18 at 15:13
• You seem to be using $A$ to represent both the blue oval and its subset the left hand crescent (similarly $B$ the red oval and the right hand crescent). This will not aid understanding – Henry Sep 15 '18 at 20:02

One of the reasons that Venn Diagrams are effective in visualizing probability problems as it poses a rather difficult concept to grasp (probability) in terms of a more intuitive concept (area). So, why don't we use area as a tool to help us prove this concept informally.

Let $P(A)$ be the area enclosed by circle $A$, and $P(B)$ the area enclosed by circle $B$. We seek to find the total area covered by both circles, i.e., $P(A\cup B)$. As you noted, $P(A\cup B)\neq P(A)+P(B)$ since the circles overlap, so we double count the area contained by both circles. How much area is double counted? The area contained by both circles, otherwise denoted by $P(A\cap B)$. This should help you develop intuition for why $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$Now, this is by no means a proper proof, but it should develop your intuition on the issue. And actually, this concept of representing probability as area is not as crazy as it may seem. As you'll learn if you continue studying probability, probability is closely related to a concept of measure, which is a representation of how much "space" a set contains.

Another way to see the proof is the following. Denote the area contained by both circles as $P(C)$. So, our equation becomes $$P(A\cup B) = P(A-C)+P(B-C)+P(C)$$Now since $C\subset A$ and $C\subset B$, $$P(A-C)=P(A)-P(C)\quad P(B-C)=P(B)-P(C)$$So, when we substitute into our equation, we get $$P(A\cup B) = P(A)+P(B)-P(C)$$

You seem to be using $A$ to represent both the blue oval and its subset the left hand crescent (similarly $B$ the red oval and the right hand crescent). This will not aid understanding

So let's instead say we have a Venn diagram with crescent $D$ on the left, intersection $E$ in the middle and crescent $F$ on the right where $D,E,F$ are disjoint events and their probabilities can be added

• $P(A)=P(D)+P(E)$
• $P(B)=P(F)+P(E)$
• $P(A \cap B)=P(E)$
• $P(A \cup B)=P(D)+P(E)+P(F)$

Now it should be reasonably clear from this final point that $$P(A \cup B)=\big(P(D)+P(E)\big)+\big(P(F)+P(E)\big)-P(E)$$ and so from the previous points that $$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$

With respect to the $3$ axioms of probability, observe that $A\cup B = (A\setminus (A\cap B)) \cup (B\setminus (A\cap B)) \cup (A\cap B)$ where the three events are disjoints, so using axiom 3 :

\begin{align*} \mathbb{P}(A\cup B) &= \mathbb{P}((A\setminus (A\cap B)) \cup (B\setminus (A\cap B)) \cup (A\cap B))\\ &= \mathbb{P}(A\setminus (A\cap B))+ \mathbb{P}(B\setminus (A\cap B)) + \mathbb{P}(A\cap B)) \end{align*}

Also observe that using axiom 3 and the fact that $A\setminus (A\cap B)$ and $A\cap B$ are disjoints \begin{align*} &\mathbb{P}(A) = \mathbb{P}(A\setminus (A\cap B)) + \mathbb{P}(A\cap B)\\ \Leftrightarrow& \mathbb{P}(A\setminus (A\cap B)) = \mathbb{P}(A) - \mathbb{P}(A\cap B) \end{align*}

Combining the two (and the same for $B\setminus (A\cap B)$), we get $$\mathbb{P}(A\cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A\cap B)$$

• I mean, sure, this is the standard proof, but I think the OP wants an intuitive understanding of the proof based on the Venn Diagram. – Don Thousand Sep 15 '18 at 15:24
• Also, this is almost exactly the proof in the link you've provided. – Don Thousand Sep 15 '18 at 15:25
• I can imagine, it's not like if there are a lot of ways of showing it. This answer is just concerning the title of the question, showing the equality. – P. Quinton Sep 15 '18 at 15:27
• It's important to understand the level of understanding of the OP in your answer. The goal is to help them, not be pedantic – Don Thousand Sep 15 '18 at 15:31
• @Rushabh Mehta Well, I think that answers are not useful only to the OP (hopefully) and since we can provide several answer it might make sense to have some that could be less intuitive. The main problem with my answer is that the proof is the same as on Wikipedia but this is a standard way of proving it. – P. Quinton Sep 16 '18 at 7:27