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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:
  1. For any event E P(E) 0
  2. P(S) = 1
  3. If E and F are mutually exclusive events, then P(E F) = P(E) + P(F).
Let S be a sample space containing the outcomes
i.e., S =

It follows from the axiomatic definition of probability that
  1. for each
  2. For any event A,
  3. where is the empty set.

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 È B) = P (A) + P (B) - P (A Ç B)

Proof
Let n be the total number of exhaustive, equally likely outcomes of the experiment. Let m1 and m2 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 m3 be the number of these common sample points. m3 will be zero in case A and B are mutually exclusive.
P (A Ç B) =

The m3 sample points which are common to both events A and B, are included in the events A and B separately.
Number of sample points in the even A È B = m1 + m2 - m3
m3 is subtracted from m1 + m2 to avoid counting of common sample points twice.
P (A È B) = = P (A) + P (B) - P (A Ç B)

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

Proof
Let n be the total number of exhaustive, equally likely cases of the experiment. Let m1 and m2 be the number of cases favorable to the happening of the events A and B respectively.
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.

The event A È B can happen in exactly m1 + m2 ways.

P (P È B) = = + = P (A) + P (B)
Hence, P (A È 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' = A = {1, 3, 5, 7, 9, 10}
Now P (A) = P(1) + P(3) + P(5) + P(7) + P(9) + P(10)
==

Thus, P (A') = = 1 - = 1 - P(A)
Also, we know that A and A are mutually exclusive and exhaustive events i.e.,
A A = φ and A A = S
or P (A A) = P (S)
Now P (A) + P (A) = 1, by using axioms (ii) and (iii).
or P( A ) = P (not A) = 1 - P (A)




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