axiom and theorem

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.