# Axiomatic Approach to Probability

Let S be a sample space of a random experiment. The probability P is a real valued function whose domain is the power set of S and range is the interval [0,1] satisfying the following axioms:

- For any event E P(E) â‰¥ 0
- P(S) = 1
- If E and F are mutually exclusive events, then P(E ïƒˆ F) = P(E) + P(F).

i.e., S =

It follows from the axiomatic definition of probability that

- for each
- For any event A,
- where is the empty set.

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# Probability of equally likely outcomes

Let S be a sample space and E be an event, such that

*n*(S) =

*n*and

*n*(E) =

*m*. If each outcome is equally likely, then it follows that

P (E) = =

# Probability of the event 'A or B'

**Theorem**

If A and B are two events not necessarily mutually exclusive, associated with a random experiment, then

P( A Ãˆ B) = P (A) + P (B) - P (A Ã‡ B)

**Proof**

Let n be the total number of exhaustive, equally likely outcomes of the experiment. Let m

_{1}and m

_{2}be the number of outcomes favorable to the happening of the events A and B respectively.

and

Since the events are given to be not necessarily mutually exclusive, there may be some sample points common to both events A and B.

Let m

_{3}be the number of these common sample points. m

_{3}will be zero in case A and B are mutually exclusive.

âˆ´ P (A Ã‡ B) =

The m

_{3}sample points which are common to both events A and B, are included in the events A and B separately.

âˆ´ Number of sample points in the even A Ãˆ B = m

_{1}+ m

_{2}- m

_{3 }

m

_{3}is subtracted from m

_{1}+ m

_{2}to avoid counting of common sample points twice.

âˆ´ P (A Ãˆ B) = = P (A) + P (B) - P (A Ã‡ B)

**Theorem**

If A and B are two mutually exclusive events associated with a random experiment, then P (A Ãˆ B) = P (A) + P (B)

**Proof**

Let n be the total number of exhaustive, equally likely cases of the experiment. Let m

_{1}and m

_{2}be the number of cases favorable to the happening of the events A and B respectively.

âˆ´ P (A) = and P (B) =

Since the events are given to be mutually exclusive, therefore, there cannot be any sample point common to both events A and B.

âˆ´ The event A Ãˆ B can happen in exactly m

_{1}+ m

_{2}ways.

âˆ´ P (P Ãˆ B) = = + = P (A) + P (B)

Hence, P (A Ãˆ B) = P (A) + P (B)

# Probability of event 'not A'

Consider the event A = {2, 4, 6, 8} associated with the experiment of drawing a card from a deck of ten cards numbered from 1 to 10. Clearly the sample space is S = {1, 2, 3, ...,10}

If all the outcomes 1, 2, ...,10 are considered to be equally likely, then the probability of each outcome is .

Now P (A) = P(2) + P(4) + P(6) + P(8)

= +++ = =

Also event 'not A' = Aâ€² = {1, 3, 5, 7, 9, 10}

Now P (Aâ€²) = P(1) + P(3) + P(5) + P(7) + P(9) + P(10)

==

Thus, P (A') = = 1 - = 1 - P(A)

Also, we know that Aâ€² and A are mutually exclusive and exhaustive events i.e.,

A âˆ© Aâ€² = Ï† and A âˆª Aâ€² = S

or P (A âˆª Aâ€²) = P (S)

Now P (A) + P (Aâ€²) = 1, by using axioms (ii) and (iii).

or P( Aâ€² ) = P (not A) = 1 - P (A)