Inclusion-Exclusion Principle: M Arton Bal Azs and B Alint T Oth October 13, 2014
Inclusion-Exclusion Principle: M Arton Bal Azs and B Alint T Oth October 13, 2014
Inclusion-Exclusion Principle: M Arton Bal Azs and B Alint T Oth October 13, 2014
This little write-up is part of important foundations of probability that were left out of the unit Probability 1
due to lack of time and prerequisites. Here we prove the general (probabilistic) version of the inclusion-exclusion
principle. Many other elementary statements about probability have been included in Probability 1. Notice
that the inclusion-exclusion principle has various formulations including those for counting in combinatorics.
We start with the version for two events:
Proof. We make use of the simple observation that E and F − E are exclusive events, and their union is E ∪ F :
On the other hand, F − E and F ∩ E are also exclusive events with union equal to F :
The difference of the two equations gives the proof of the statement.
Next, the general version for n events:
P{E1 ∪ E2 ∪ · · · ∪ En }
X X X
= P{Ei }− P{Ei1 ∩Ei2 }+ P{Ei1 ∩Ei2 ∩Ei3 }−· · ·+(−1)n+1 P{E1 ∩E2 ∩· · ·∩En }.
1≤i≤n 1≤i1 <i2 ≤n 1≤i1 <i2 <i3 ≤n
Intuitively, summing the probabilities we double-count all the two-intersections. Those we subtract with the
second sum. (Observe that every two-intersection is contained exactly once in {Ei1 ∩ Ei2 : 1 ≤ i1 < i2 ≤ n}.)
Unfortunately, with this move we have now counted all three-intersections three times, then subtracted them
three times, hence we have to add them back once. But then we run into trouble with four-intersections, etc.
When our state space is countable then counting arguments give a direct proof of the formula. This can also
be extended to the general case. Here we give a different proof.
Proof. We argue inductively. The proof for n = 2 is seen above. Suppose that the formula is true for n, we
show it for n + 1. First apply the n = 2 case, then distributivity of intersections:
P{E1 ∪ E2 ∪ · · · ∪ En ∪ En+1 }
= P{(E1 ∪ E2 ∪ · · · ∪ En ) ∪ En+1 }
= P{E1 ∪ E2 ∪ · · · ∪ En } + P{En+1 } − P{(E1 ∪ E2 ∪ · · · ∪ En ) ∩ En+1 }
= P{E1 ∪ E2 ∪ · · · ∪ En } + P{En+1 } − P{(E1 ∩ En+1 ) ∪ (E2 ∩ En+1 ) ∪ · · · ∪ (En ∩ En+1 )}.
∗ University of Bristol / Budapest University of Technology and Economics
1
The first and the last terms are n-unions, for which we assumed the formula to hold. Therefore
X
P{E1 ∪ E2 ∪ · · · ∪ En ∪ En+1 } = P{Ei } (1)
1≤i≤n
X
− P{Ei1 ∩ Ei2 } (2)
1≤i1 <i2 ≤n
X
+ P{Ei1 ∩ Ei2 ∩ Ei3 } (3)
1≤i1 <i2 <i3 ≤n
Here (1) and (5) account for all the probabilities of single events from 1 to n + 1. (2) includes all the two-
intersection probabilities from 1 to n, and (6) all the two-intersection probabilities where the higher index equals
n + 1. These two sums thus account for all possible two-intersection probabilities from 1 to n + 1. Similarly,
(3) includes all three-intersection probabilities from 1 to n, and (7) those with highest index equal to n + 1.
Together they include all three-intersection probabilities from 1 to n + 1. This continues until (4) and (8), which
together give all n-intersection probabilities from 1 to n + 1. Finally, we write down the last term, and
P{E1 ∪ E2 ∪ · · · ∪ En+1 } =
X X X
= P{Ei } − P{Ei1 ∩ Ei2 } + P{Ei1 ∩ Ei2 ∩ Ei3 }
1≤i≤n+1 1≤i1 <i2 ≤n+1 1≤i1 <i2 <i3 ≤n+1
X
− · · · + (−1)n+1 P{Ei1 ∩ Ei2 ∩ · · · ∩ Ein } + (−1)n+2 P{E1 ∩ E2 ∩ · · · ∩ En+1 },
1≤i1 <i2 <···<in ≤n+1
Corollary 3 The right hand-side of the inclusion-exclusion formula alternates in the sense that the first sum
is greater than or equal to the probability of the union on the left hand-side. The difference of the first two sums
is smaller than or equal to the left hand-side. The first three sums together with their signs are larger than or
equal, etc.
Proof. This statement can be followed in an inductive fashion along the proof.