Lecture 2
Properties of Probability.
Combinatorics
Text: Sections 2.3, 2.4, 3.1, 3.2
Print version of the lecture in MA340 Probability Theory presented on September
13, 2012
2.1
Contents
1 Properties of Probability 1
2 Law of Partitions 3
3 Specifying Probabilities 5
3.1 Countable Probability Space . . . . . . . . . . . . . . . . . . 5 . . . . . .
3.2 Classical Probability . . . . . . . . . . . . . . . . . . . . . 5 . . . . . . .
3.3 Geometric Probability . . . . . . . . . . . . . . . . . . . . . 6 . . . . . . .
4 Buffons Needle Problem 6
5 Counting Methods 7
6 Combinatorial Coefcients 10 2.2
1 Properties of Probability
Axioms of Probability
The probability measure P dened on the set of events of W satises the following three
axioms.
Axiom 1 (Nonnegativity) P(E) 0
Axiom 2 (Certainty) P(W) = 1
Axiom 3 (Countable Additivity) For each countable collection of pairwise disjoint events
E1;E2;::: s.n.mE E/ when n 6= m we have that
[ !
P En = P(En)
n1 n1
2.3
1 Properties of Probability (13)
1. P(0/) = 0 (Probability of an impossible event)
Proof: 0/ = W ;0/ [W = W;0 / \W = 0 /. Hence, by exclusivity of the two sets/ and
W, and by and axiom 3, we have:
P(W) = P(0 / [W) = P(0/)+P(W) =) P(0 /) = 0:
2. P(A ) = 1P(A) (Complementation Rule)
Proof:Write W as a disjoint union, W = A[A . Then, by axioms 2 and 3:
1 = P(W) = P(A)+P(A ): c
3. A B ) P(A) P(B) (Domination Principle)
Proof: Since A[A = W, we can write B = B\W = (B\A)[(B\A ), i.e. as a c
union of two mutually exclusive events. Hence,
P(B) = P(B\A)+P(B\A ) = P(A)+P(B\A ) c
where we used the fact that A B =) B\A = A. Then P(B) P(A) since the event
E B\A W has probability P(E) 0 by axiom 1.
2.4
Properties of Probability (45)
4. 8A W P(A) 1
Proof: By property 3 above and axiom 2, P(A) P(W) = 1.
5. P(A[B) = P(A)+P(B)P(AB) (Addition Rule)
Proof: Write A[B as union of 3 disjoint sets:
A[B = (An(A\B))[(A\B)[(Bn(A\B))
and use P(E nF) = P(E)P(F) with axiom 3 to obtain
P(A[B) = P(A)P(A\B)+P(A\B)+P(B)P(A\B)
= P(A)+P(B)P(A\B)
2.5
2Properties of Probability (6)
6. 8A 1:::;A N W we have the Inclusion-Exclusion Identity:
N ! N
[
P A k = P(A k P(A k1)k2
k=1 k=1 k1

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