Visually explaining this probability union rule

Question

I'm trying to visually wrap my head around the following equivalence of the union probability rule: $$ P(A)+P(B)-P(A)P(B)=P(A)+(1-P(A))P(B) $$

I understand that $(1-P(A))=P(A')$ and the whole equivalence makes sense to me algabraically.

However, I have sketched these Venn diagrams to try and visualise the equivalence, and what I am struggling to understand is that if I was to combine the Venn diagrams for $P(A)$ and $P(A')P(B)$ then I would have $P(A)+P(B)$, not $P(A)+P(B)-P(A)P(B)$.

Where am I tripping up in my reasoning?

Answer 1

$P(A)$ is the probability of $A$ occurring, and $P(B)$ is the probability of $B$ occurring.

The probability of either $A$ or $B$ occurring is $P(A \cup B)$.

But $P(A \cup B)$ does not in general equal $P(A) + P(B)$.

If you look in your diagrams you see that $P(A) + P(B)$ counts $P(A \cap B)$ twice.

Answer 2

You can't use visual diagrams to show these kinds of numerical identities the way you do. To be precise, with your visual method of looking at certain regions as unions, intersections, or complements of more basic regions, and translating that directly into numerical operations of addition, multiplication, and subtraction from one, you would get that $P(A \cup B) = P(A) + P(B)$ and $P(A \cap B) = P(A)P(B)$, but that's not true in general.

As others have pointed out: $P(A\cup B)$ equals $P(A)+P(B)$ only when $A$ and $B$ are mutually exclusive events. If not, you end up double-counting. As the most basic example example, your method gets the result that $P(A)=P(A\cup A) =P(A) + P(A)$ which is of course not true.

Now, as also pointed out, you could still use a visual method and handle such double-counting by using different shades of color in the same region.

However, this does not fix the even more problematic implicit assumption in your approach that $P(A \cap B) =P(A)P(B)$, which is only true if $A$ and $B$ are independent. Again, as a trivial case where they are not, your method would say that $P(A) =P(A\cap A)=P(A)P(A)$ which again is clearly not true.

And, for this problem, there is no visual bandaid: you really can't use a qualitative relationship between events to depict their quantitative dependency. Or, more to the point: you can't infer the qualitative formula from the visual depiction.

Answer 3

correction: $P\Big(A\cup B\Big)\equiv P(A)+P(B)-P(A\cap B)$

correction: $P\Big(A\cap B\Big)\not\equiv P(A)\,P(B)$

correction: $P\Big((A\cup B)\setminus(A\cap B)\Big)\equiv P(A)+P(B)-2\, P(A\cap B)$

$P\Big(A'\cap B\Big)\equiv P\Big(B\setminus A\Big)\equiv P(B)-P(A\cap B)\not\equiv P(A')\,P(B)$

   Note:

1. $\large {P\setminus Q}$ (read “$P$ minus $Q$”) means to subtract $P\cap Q$ from $P.$

   $\large {P\setminus Q}\:$ is also written as $\:\large{P-Q};$ the former is a better idea, to discourage conflating set operations and arithmetic operations (how you were conflating $\cup$ and $+$ and then $\cap$ and $\times$).

2. $P(A\cap B)=P(A)\,P(B)\,$ if and only if events $A$ and $B$ are independent.