Feedback on attempt at solution to $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A\cap C) - P(A\cap B) -\cdots$

Question

Verify: $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A\cap C) - P(A\cap B) - P(A\cap C) - P(B \cap C) + P(A \cap B \cap C)$

I have looked around on the site and seen solutions for the question, but what I was looking for here is feedback on where I went wrong in my attempt at the solution and is there a way for me to reconcile what I did with the solutions that I have seen here?

Attempt

1) Decomposed $A \cup B \cup C$ into a collection of disjoint subsets:

$$D = A \cap B^{c} \cap C^{c} \\ E = A^{c} \cap B \cap C^{c} \\ F = A^{c} \cap B^{c} \cap C \\ G = A \cap B \cap C^{c} \\ H = A \cap B^{c} \cap C \\ I = A^{c} \cap B \cap C \\ J = A \cap B \cap C$$

2) By the fundamental axioms of probability:

$$P(A \cup B \cup C) = P(D) + P(E) + P(F) + P(G) + P(H) + P(I) + P(J)$$

Observe that:

$$ A = D \cup G \cup H \cup J \\ B = E \cup G \cup I \cup J \\ C = F \cup H \cup I \cup J$$

Therefore we can say

$$P(A) + P(B) + P(C) = P(D) + P(E) + P(F) + P(G) + P(H) + P(I) + P(J) + P(E) + P(G) + P(I) + P(J) + P(F) + P(H) + P(I) + P(J) \\ \Rightarrow P(A) + P(B) + P(C) = P(A \cup B \cup C) + P(G) + P(H) + P(I) + 2P(J) $$

Rearranging terms then yields:

$$ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(G) - P(H) - P(I) - 2P(J) \\ = P(A) + P(B) + P(C) - P(A \cap B) - P(A\cap C ) - P(B \cap C) - 2P(A \cap B \cap C)$$

As can be seen, everything works out except for that last term....So where did I go wrong in my approach? I feel I didn't treat an intersection of sets in the correct way.

Answer 1

At a glance, it looks like your mistake is on the last line.

$P(A\cup B\cup C)=P(A)+P(B)+P(C)\color{red}{-P(G)}\color{blue}{-P(H)}\color{green}{-P(I)}-2P(J)$

$=P(A)+P(B)+P(C)\color{red}{-P(A\cap B)}\color{blue}{-P(A\cap C)}\color{green}{-P(B\cap C)}-2P(A\cap B\cap C)$

It appears that you thought that $P(G)$ was equal to $P(A\cap B)$ and similarly with the others, but that is not quite right. Instead $P(A\cap B)=P(G)+P(J)$

So, adding and subtracting $P(J)$ (which is equivalent to having added zero and is perfectly fine to do) for each of these we get:

$P(A\cup B\cup C)=P(A)+P(B)+P(C)\color{red}{-P(G)}\color{blue}{-P(H)}\color{green}{-P(I)}-2P(J)\\\underbrace{\color{red}{-P(J)}+P(J)}_{=0}\color{blue}{-P(J)}+P(J)\color{green}{-P(J)}+P(J)$

Using both red terms for $P(A\cap B)$ and similarly for the others, and seeing $-2P(J)+3P(J)$ for a total of $+1P(J)$ and replacing, this gives:

$=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$

Answer 2

It might be less of a confusion to break it down into two steps.

Firstly show for any two events $X$, $Y$ that: $$\small\begin{align}\mathsf P(X\cup Y)&=\mathsf P(X\smallsetminus Y)+\mathsf P(X\cap Y)+\mathsf P(Y\smallsetminus X)\\&~~\vdots\\&=\mathsf P(X)+\mathsf P(Y)-\mathsf P(X\cap Y)\end{align}$$

So now you can extend this to any three events $A$, $B$, and $C$:$$\small\begin{align}\mathsf P(A\cup B\cup C)&=\mathsf P(A)+\mathsf P(B\cup C)-\mathsf P(A\cap (B\cup C))\\&=\mathsf P(A)+\mathsf P(B\cup C)-\mathsf P((A\cap B)\cup(A\cap C))\\&~~\vdots\\&=\mathsf P(A)+\mathsf P(B)+\mathsf P(C)-\mathsf P(A\cap B)-\mathsf P(A\cap C)-\mathsf P(B\cap C)+\mathsf P(A\cap B\cap C)\end{align}$$