This document defines probability and describes the axioms and theorems related to probability functions. It provides examples to illustrate key concepts. Specifically, it defines probability as the number of outcomes in an event divided by the total number of outcomes in the sample space. It then outlines three axioms for probability functions: 1) the probability of an event is between 0 and 1, 2) the probability of the sample space is 1, and 3) the probability of the union of mutually exclusive events equals the sum of their individual probabilities. Several theorems are also presented, such as the probability of the complement of an event equals 1 minus the probability of the event.

1. Probability P is a real valued function that assigns to each
event E in the sample space S a number P(E)
A probability of an event E is given by
P(E) = N(E) / N(S)
i.e. (No of outcomes in E) / (No of outcomes in S)

2. Example
Experiment : Throwing a coin twice
Define events A and B as follows
A = { At least one H is obtained } B= { at least one T is
obtained}
= { HT, TH, HH} = { TH, HT, TT}
S= { HT,HH,TH,TT}
Then N(S) = 4, N(A) = 3, N(B) = 3
a) P(A) = 3/4
b) P(A ∩ B) = 2/4
c) P(AC) = 1/4

3. Axioms for Probability Functions
Axiom 1 : For every event A in a sample space S, 1 ≥ P(A) ≥ 0
Axiom 2 : P(S) = 1
Axiom 3 : Let A and B be any two mutually exclusive events
defined over S. Then
P(A U B) = P(A) + P(B)

4. Example:
Consider a random trial that can result in failure or success. Let 0 stand for failure, and
let 1 stand for success. Then we can consider the outcome space to be S = {0, 1}. For
any number p between 0 and 100%, define the function P as follows:
P({1}) = p,
P({0}) = 100% − p,
P(S) = 100%,
P({}) = 0.
Then P is a probability distribution on S, as we can verify by checking that it satisfies
the axioms:
1. Because p is between 0 and 100%, so is 100% − p. The outcome space S has but
four subsets: {}, {0}, {1}, and {0, 1}. The values assigned to them by P are 0, 1 − p,
p, and 100%, respectively. All these numbers are at zero or larger, so P satisfies
Axiom 1.
2. By definition, P(S) = 100%, so P satisfies Axiom 2.
3. The empty set and any other set are disjoint, and it is easy to see that
P({}∪A) = P({}) + P(A) for any subset A of S.
The only other pair of disjoint events in S is {0} and {1}. We can calculate
P({0}∪{1}) = P(S) = 100% = (100% − p) + p = P({0}) + P({1}).
Thus P satisfies Axiom 3.

5. Theorem for Probability Functions
Theorem 1
If Ac is a complement of A , then
P(Ac) = 1- P(A)
Proof :
S = Ac A,
from Axiom 2 and 3, P(S) = 1 = P(A) + P(Ac) ,
since A and Ac are ME
P(Ac) = 1 – P(A)

6. Theorem 2
P(Ø) = 0, the impossible event has probability
zero.
Proof :
Let S = Ac A, where A = Ø
Then
S = Ac Ø Ac = S

7. Theorem 3
If A B, then P(A) ≤ P(B)
Proof :
We can write B as B = A (B A’) where A and
(B A’) are ME.
Thus, P(B) = P(A) + P(B A’)
But P(B A’) 0 since A B
This P(B) P(A)

8. Theorem 4
For any event A, P(A) ≤ 1
Proof :
We know that A S and P(S) = 1 P(A) ≤ P(S)
= 1

9. Theorem 5
If A and B are any two events, then
P(A B) = P(A) +P(B) - P(A B)
Proof :
A B=A ( Ac B)
P(A B) = P(A) + P( Ac B)
P( Ac B) = P(A B) - P(A) ……………….. (1)
But
B = (A B) (Ac B)
P(B) = P(A B) + P(Ac B)
P( Ac B) = P(B) - P(A B) …………… (2)
(1) = (2)
P(A B) - P(A) = P(B) - P(A B)
P(A B) = P(A) + P(B) - P(A B)

10. Thm 6
If A , B and C are any three events, then
P(A B) = P(A) +P(B) - P(A B)
Proof :
P(A) = P(A Bc) + P (A B) and
P(B) = P(B Ac) + P (A B)
adding these two equation
P(A) + P(B) = [P(A Bc) + P(B Ac) + P (A B)] + P (A B)
the sum in the bracket is P(A B).If we subtract P (A B) both
sides of the equation,the result follws.