Axioms of Probability

In this class, We discuss Axioms of Probability.

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The reader should have prior knowledge of probability basics. Click Here.

Axiom 1: For any event, A

P(A) >= 0

The event which is unlikely to happen P(E) ≈ 0

Example: E = Randomly select card ‘5’ of spade

P(E) = 1/52

The event likely to happen P(E) ≈ 1

Example: E = Randomly pick spade or red card.

P(E) = 39/52

Axiom 2:

Probability of sample space S = 1

Definitely one of the possibility from the sample space will happen.

Example: Toss two coins

S = {HH, HT, TH, TT}

P(S) = 1

100% one from the sample space will happen

Axiom 3:

If A1, A2, A3, . . . An are disjoint events.

P(A1 U A2 U . . An) = P(A1) + P(A2) + . . P(An)

Example: Toss two coins

S = {HH, HT, TH, TT}

E1 = {HH}

E2 = {HT}

E1 and E2 are disjoint events.

P(E1 U E2) = P(E1) + P(E2)

P(E1 U E2) = 1/4 + 1/4 = 1/2

Without using the Axiom

E = {HH, HT}

P(E) = 2/4 = 1/2

Not Disjoint Events

Addition law of probability

P(E1 U E2) = P(E1) + P(E2) – P(E1 ∩ E2)

Example: Roll a dice

E1 = dice show even number = {2, 4, 6}

E2 = dice show value greater than three = {4, 5, 6}

The below Venn diagram of events E1 and E2 are not disjoint sets.

Axioms of Probability 1
Not Disjoint Sets

P(E1) = 3/6

P(E2) = 3/6

P(E1 ∩ E2) = 2/6

The numbers 4 and 6 are involved in the probability of events E1 and E2.

One time is enough. So we are subtracting P(E1 ∩ E2)

Similarly, P(E1 U E2 U E3) = P(E1) + P(E2) + P(E3) – P(E1 ∩ E2) – P(E1 ∩ E3) – P(E2 ∩ E3) + P(E1 ∩ E2 ∩ E3)

Example: Roll a dice

A = {1, 2, 5}

B = {2, 5, 6}

C = {5, 6, 3}

the common element is 5

P(A) = 3/6 the element 5 involved

P(B) = 3/6 the element 5 involved

P(C) = 3/6 the element 5 involved

P(A ∩ B) = 2/6 the element 5 involved

P(A ∩ C) = 1/6 the element 5 involved

P(B ∩ C) = 2/6 the element 5 involved

P(A ∩ B ∩ C) = 1/6

P(A U B U C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)

We added the term P(A ∩ B ∩ C) to involve element five.

Complement Law:

P(E’) = 1 – P(E)

P(E) + P(E’) = 1

P(E) + P(E’) = P(S)