Probability of general celebration in class
● Buses are delayed/cancelled 10% of the time during the winter
● I oversleep 20% of the time
Modeling
● A: event that buses are delayed– (or frst component breaks)
● B: event that I oversleep– (or second component breaks)
● Late = A ∪ B: event that I am late– (or current is blocked)
Probability of a Union
● Kolmogorov's 3rd Axiom guarantees a simple formula for the probability of the union of mutually exclusive events in a probability space
P(E1 ∪ E
2 ∪ E
3 ∪ …) = P(E
1) + P(E
2) + P(E
3) + …
Probability of a Union
● Kolmogorov's 3rd Axiom guarantees a simple formula for the probability of the union of mutually exclusive events in a probability space
P(E1 ∪ E
2 ∪ E
3 ∪ …) = P(E
1) + P(E
2) + P(E
3) + …
● But what if the events are not mutually exclusive?
Counting Elements
|A ∪ B| = |A ∪ (B \ A)|
= |A| + |B \ A|
= |A| + |B \ A| + |A ∩ B| – |A ∩ B|
= |A| + |(B \ A) ∪ (A ∩ B)| – |A ∩ B|
Counting Elements
|A ∪ B| = |A ∪ (B \ A)|
= |A| + |B \ A|
= |A| + |B \ A| + |A ∩ B| – |A ∩ B|
= |A| + |(B \ A) ∪ (A ∩ B)| – |A ∩ B|
= |A| + |B| – |A ∩ B|
Counting Elements
|A ∪ B| = |A ∪ (B \ A)|
= |A| + |B \ A|
= |A| + |B \ A| + |A ∩ B| – |A ∩ B|
= |A| + |(B \ A) ∪ (A ∩ B)| – |A ∩ B|
= |A| + |B| – |A ∩ B|
A similar result holds for probabilities
The Inclusion-Exclusion Principle(for two events)
For two events A, B in a probability space:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
The Inclusion-Exclusion Principle(for two events)
For two events A, B in a probability space:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Don't use this to “prove” Kolmogorov's Axioms!!!
Proof:
P(A ∪ B) = P(A ∪ (B \ A)) (set theory)
= P(A) + P(B \ A) (mut. excl., so Axiom 3)
The Inclusion-Exclusion Principle(for two events)
Proof:
P(A ∪ B) = P(A ∪ (B \ A)) (set theory)
= P(A) + P(B \ A) (mut. excl., so Axiom 3)
= P(A) + P(B \ A) + P(A ∩ B) – P(A ∩ B) (Adding 0 = P(A ∩ B) – P(A ∩ B) )
The Inclusion-Exclusion Principle(for two events)
Proof:
P(A ∪ B) = P(A ∪ (B \ A)) (set theory)
= P(A) + P(B \ A) (mut. excl., so Axiom 3)
= P(A) + P(B \ A) + P(A ∩ B) – P(A ∩ B) (Adding 0 = P(A ∩ B) – P(A ∩ B) )
= P(A) + P((B \ A) ∪ (A ∩ B)) – P(A ∩ B) (mut. excl., so Axiom 3)
The Inclusion-Exclusion Principle(for two events)
Proof:
P(A ∪ B) = P(A ∪ (B \ A)) (set theory)
= P(A) + P(B \ A) (mut. excl., so Axiom 3)
= P(A) + P(B \ A) + P(A ∩ B) – P(A ∩ B) (Adding 0 = P(A ∩ B) – P(A ∩ B) )
= P(A) + P((B \ A) ∪ (A ∩ B)) – P(A ∩ B) (mut. excl., so Axiom 3)
= P(A) + P(B) – P(A ∩ B) (set theory)
The Inclusion-Exclusion Principle(for two events)
Will I be late for class?
● P(Late) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= 0.1 + 0.2 – ???
Let's make the modeling assumptionA and B are independent: P(A ∩ B) = P(A) P(B)
Will I be late for class?
● P(Late) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – P(A) P(B)
= 0.1 + 0.2 – 0.1 × 0.2
Will I be late for class?
● P(Late) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – P(A) P(B)
= 0.1 + 0.2 – 0.1 × 0.2
= 0.28
The Inclusion-Exclusion Principle(for three events)
For three events A, B, C in a probability space:
P(A ∪ B ∪ C) = P(A) + P(B) + P(C)
– P(A ∩ B) – P(B ∩ C) – P(C ∩ A)
+ P(A ∩ B ∩ C)
The Inclusion-Exclusion Principle
For events A1, A
2, A
3, … A
n in a probability space:
P(∪i=1n Ai)=∑i=1
n
P(A i)−∑1≤i< j≤nP(Ai∩A j)
+∑1≤i< j<k≤nP(Ai∩A j∩A k)−...+(−1)n−1P(∩i=1
n A i)