Solutions to Proof Assignment #1
1.(3 points) Prove: If A and B are subsets of Ω, then
P (A ∪ B) = P (A) + P (B) − P (A ∩ B).
(Note: This exercise is proven in our textbook, but the proof is written poorly. Please write
the proof in your own words and make sure that you justify every statement that you make.
You will receive NO CREDIT if you simply copy the proof from the book.)
If A, B are disjoint subsets of Ω then by property (4) of Theorem 1 (from class), we know
that P (A ∪ B) = P (A) + P (B). Thus, assume that A, B are not disjoint. From the lecture
on set theory, we know that we can write A, B as a union of disjoint sets in the following
manner:
A = (A ∩ B c ) ∪ (A ∩ B)
B = (B ∩ Ac ) ∪ (B ∩ A)
Then, P (A) = P (A ∩ B c ) + P (A ∩ B) and P (B) = P (B ∩ Ac ) + P (B ∩ A), by property
(4) of Theorem 1. Adding these together, we obtain:
P (A) + P (B) =
=
=
[P (A ∩ B c ) + P (A ∩ B)] + [P (B ∩ Ac ) + P (B ∩ A)]
P (A ∩ B c ) + P (B ∩ Ac ) + 2P (A ∩ B)
X
X
X
m(ω) +
m(ω) + 2
P (ω),
ω∈A∩B c
ω∈B∩Ac
ω∈A∩B
where the last equality follows from the definitions of P (A ∩ B), P (A ∩ B c ) and P (B ∩ Ac ).
However, from the definition of P (A ∪ B), we see that
X
X
X
X
X
m(ω) = P (A∪B)+P (A∩B).
m(ω)+
m(ω) =
m(ω)+2
m(ω)+
ω∈A∩B c
ω∈B∩Ac
ω∈A∩B
ω∈A∪B
ω∈A∩B
Therefore, by subtracting P (A ∩ B) from both sides of the equation above, we obtain
P (A ∪ B) = P (A) + P (B) − P (A ∩ B).
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Solutions to Proof Assignment #1
2.(2 points) What do you think the formula should be for P (A ∪ B ∪ C) where A, B, C are
subsets of Ω (not necessarily pairwise disjoint)? Explain your answer using a Venn diagram.
The formula should be:
P (A ∪ B ∪ C) = P (A) + P (B) + P (C) − P (A ∩ B) − P (B ∩ C) − P (A ∩ C) + P (A ∩ B ∩ C).
If you draw a Venn Diagram, you’ll notice that when we sum P (A) + P (B) + P (C),
we are adding each of P (A ∩ B), P (B ∩ C) and P (A ∩ C) twice. So, we subtract out
P (A ∩ B), P (B ∩ C) and P (A ∩ C) so that they aren’t being double-counted.
But, now we run into another problem: By adding P (A) + P (B) + P (C), we’ve added
P (A ∩ B ∩ C) three times. Moreover, by subtracting P (A ∩ B), P (B ∩ C) and P (A ∩ C),
we’ve subtracted P (A ∩ B ∩ C) three times, giving us a net result of 0 copies of P (A ∩ B ∩ C)
in our sum. So, we need to add it back in so that P (A ∩ B ∩ C) is counted once.
Note: This idea of alternating between subtracting and adding to avoid double counting
overlapping sets is called the Principle of Inclusion-Exclusion.
Bonus (+2 points) Prove: If we remove two opposite corners from a chessboard, the board
cannot be covered by dominoes (each domino covers two neighboring squares on the chessboard).
(Hint: Your answer should be clear and convincing with no cases to distinguish. If done
correctly, your answer should take no more than 2 or 3 sentences.)
A domino lies on two adjacent squares, so it must lie on one square of each color. However,
if you delete two opposite corners of a chessboard, both squares that you delete will have
the same color. Since # Black Squares 6= # White Squares, then it is impossible to arrange
the dominoes on the board.
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