This definition was given in 1930’s by Van Mises and A Kolmogorov. The limitation of both the definitions of probability given earlier are overcome by this approach to probability. The classical and relative frequency approach are special cases of this definition. It is based on the concepts of set theory.
Hence, to understand this approach, the concepts of set theory is essential.

Let  be a sample space and let p be a real-valued function defined on the subsets of . Then P is called probability function and the probability of the occurrence of event A is given by P(A) if the following axioms hold good.
Definition: Let (, S) be a sample space. A set function P defined on S is called a probability measure (or simply probability) if it satisfies the following conditions

Axiom 1: P(A) is a real number, i.e.,
P(A)  0 for every A  S or 0 P(A) 1.

Axiom 2: P() = 1.
Axiom 3: Let {Ai}, Ai  S i=1, 2, 3, …, be a disjoint sequence of sets, such that Ai  Aj =  for ij, then
We call P(A) the probability of an event. The axiom 3 is called countable additivity.
In other words, if the likelihood of the occurrence of an event A resulting from a statistical experiment is evaluated by means of a set of real numbers called weights or probabilities ranging from 0 to 1, then the probability of A is the sum of the weights of all the sample points in A.
Thus, if  is a sample space consisting of n events and if each event is equally likely to happen, then its probability is 1/n. Now if A is an event of  containing m elements of , i.e., n(A) = m, then
P(A) = 1/n + 1/n + 1/n +…+ up to m
terms
= m/n
P(A) = n(A)/n()
Where,
n(A) = No. of distinct elements in A
n() = No. of distinct elements in 

Example: If a coin is tossed thrice. What is the probability that at least one head occurs?
Solution:
 = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
If the coin is unbiased, each of these outcomes would be equally likely to occur. A probability or weight () can be assigned to each sample point. Then 8 = 1 or =1/8. If A denotes the event of at least one head occurring then, A = {TTH, THT, HTT, THH, HTH, HHT, HHH}
and
P(A) = 1/8+1/8+1/8+1/8+1/8+1/8+1/8
= 7/8

Example 2: If a coin is tossed thrice. Find the probability of getting (i) exactly one head (ii) exactly two heads, (iii) exactly one head or two heads, (iv) none of them are head
Theorem 1:
The probability of an impossible event is zero, i.e., P() = 0
Proof:
We know that, an impossible event contains no sample point, thus, the certain event  and the impossible event  are mutually exclusive.
It follows that
   = 
P(  ) = P()
P() + P() = P() from axiom 3
P() = 1-1 = 0 from axiom 1
P() = 0

Theorem 2: If Ac is the complementary event of A, the P(Ac) = 1 – P(A)

Proof:
Since A and Ac are complementary events, they are disjoint events. Also,
A  Ac = 
P(A  Ac) = P(A) + P(Ac) = P() =1
 P(Ac) = 1 – P(A)
Theorem 3:
For any two events A and B
P(Ac B) = P(B) - P(A  B)

Proof:
It is obvious that Ac A and A  B are disjoint events and
(A  B)  (Ac B) = B
Thus using axiom 3 we get,
P(B) = P(A  B) + P(Ac B)
 P(Ac B) = P(B) - P(A  B).

Similarly, it can be proved that
P(A  Bc) = P(A) - P(A  B).
Proof:
(i) Observe that A  Bc and B are
mutually exclusive events and
B  (A  Bc) = A, since B  A

 P(A) = P[B + P(A Bc)]
Using axiom 3, we get
= P(B) + P(A  Bc)
 P(A  Bc) = P(A) – P(B)

Proof:
We have B=(Ac B)  (A  B)
Now,
P(Ac B) = P(B) –P(AB) (1)
and
P(A  Bc) = P(A) – P(A  B) (2)
Adding (1) and (2), we get
P(ABc)+P(AcB)=P(A)+P(B)-2P(AB)
or
P(ABc)P(AcB)=P(A)+P(B)-2P(AB)

Laws of probability

Theorem 6:
If A and B are any two events, which are not mutually exclusive, then
P(A B) = P(A) + P(B) – P(A  B)

Proof:
We know that
A  B = A  (Ac  B)

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