Show that P(A1 U A2 U A3) = P(A1) + P(A2) + P(A3) - P(A1 n A2) - P(A2 n A3) - P(A1 n A3) + P( A1 n A2 n A3)

I could do this by drawing a few ven diagrams but is there a specific proof for it?

2. Originally Posted by shtaM
Show that P(A1 U A2 U A3) = P(A1) + P(A2) + P(A3) - P(A1 n A2) - P(A2 n A3) - P(A1 n A3) + P( A1 n A2 n A3)

I could do this by drawing a few ven diagrams but is there a specific proof for it?
Hi shtaM,

Are you familiar with the rule

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$?

If so, write

$A1 \cup A2 \cup A3 = (A1 \cup A2) \cup A3$

and see if you can apply the rule above to find

$P((A1 \cup A2) \cup A3)$.

3. I can offer 2 proofs:

1) This problem is equivalent to proving:

$
|A \cup B \cup C|=|A|+|B|+|C|-|A \cap B|-|A \cap C|-|B \cap C|+|A \cap B \cap C|
$

So we can verify that every element in (A U B U C) gets counted exactly once.

Take any a in A such that a is not in B or C, then a gets counted once in |A| and that's it because it doesn't appear in any intersection. The argument is the same for B and C. So any element that belongs to only one set is counted exactly once.

Take any x in A and B but not in C. This element gets appears once in A, once in B and once in (A n B) (1+1-1=1) so x gets counted exactly once. Same logic applies to (A n C) and (B n C). So any element that appears in exactly one intersection gets counted exactly once.

Take any y in A, B and C. Then this element appears in every term so we have 1+1+1-1-1-1+1=1. So y gets counted only once.

So every element gets counted exactly once, thus the equality holds.

2) The other one is using the rule awkward just said. Good luck.