Axiomatic Approach to Probability

Axiomatic approach is another way of describing probability of an event. In this approach some axioms or rules are depicted to assign probabilities. 
It follows from the axiomatic definition of probability that 
(i) `0 ≤ P (ω_i) ≤ 1 "for each"  ω_i ∈ S `
(ii) `P (ω_1) + P (ω_2) + ... + P (ω_n)` = 1 
(iii) For any event A, `P(A) = ∑ P(ω_i ), ω_i ∈ A.` 
For example, in ‘a coin tossing’ experiment we can assign the number `1/2` to each of the outcomes H and T.
i.e.  P(H) = `1/2` and  and P(T) = `1/2`   ...(1)
The both the conditions i.e., each number is neither less than zero nor greater than 1 and 
P(H) + P(T) = `1/2 +1/2` = 1
Therefore, in this case we can say that probability of H = `1/2` , and  probability of
T = `1/2`
If we take P(H) = `1/4` and P(T) = 3/4      ...(2)
We find that both the assignments (1) and (2) are valid for probability of H and T. In fact, we can assign the numbers p and (1 – p) to both the outcomes such that 0 ≤ p ≤ 1 and P(H) + P(T) = p + (1 –  p) = 1 
This assignment, too, satisfies both conditions of the axiomatic approach of probability.

1)   Probability of an event :
Let S be a sample space associated with the experiment ‘examining three consecutive pens produced by a machine and classified as Good (non-defective) and bad (defective)’. 
For example:
A sample space associated with this experiment is 
S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}, 
where B stands for a defective or bad pen and G for a non – defective or good pen. 
Let the probabilities assigned to the outcomes be as follows 

Let event A: there is exactly one defective pen and event B: there are atleast two defective pens. 
Hence A = {BGG, GBG, GGB} and
B = {BBG, BGB, GBB, BBB}
Now P(A) =  P(BGG) + P(GBG) + P(GGB) = `1/8 + 1/8 +1/8 = 3/8`
and P(B) =  P(BBG) + P(BGB) + P(GBB) + P(BBB) = `1/8+1/8+1/8+1/8 = 4/8 = 1/2`  

2)  Probabilities of equally likely outcomes:
Let a sample space of an experiment be 
`S = {ω_1, ω_2,..., ω_n}.` 
Let all the outcomes are equally likely to occur, i.e., the chance of occurrence of each simple event must be same. i.e. 
`P(ω_i)` = p, for all `ω_i` ∈ S where 0 ≤  p ≤ 1 
Since  \[\displaystyle\sum_{i=1}^{n} P(\omega_i) \]  = 1
i.e., p + p +...+ p (n times) = 1
or np = 1 i.e., p = 1/n
Let S be a sample space and E be an event, such that n(S) = n and n(E) = m. If each out come is equally likely, then it follows that
P(E) = `m/n = ("Numberof outcomes favourable to E")/ ("Total possible outcomes")` 

3)  Probability of the event ‘A or B’:
 Find the probability of event ‘A or B’, i.e., P (A ∪ B)
Let A = {HHT, HTH, THH} and B = {HTH, THH, HHH} be two events associated with ‘tossing of a coin thrice’ 
Clearly A ∪ B = {HHT, HTH, THH, HHH} 
Now P (A ∪ B)
= P(HHT) + P(HTH) + P(THH) + P(HHH)
If all the outcomes are equally likely, then

`P(A U B) = 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2`

Also P(A) = P(HHT) + P(HTH) + P(THH) =  `3/8`

and P(B) = P(HTH) + P(THH) + P(HHH) = `3/8`

Therefore P(A) + P(B) =` 3/8 + 3/8 = 6/8`

It is clear that P(A∪ B) ≠ P(A) + P(B) 

The points HTH and THH are common to both A and B. In the computation of P(A) + P(B) the probabilities of points HTH and THH, i.e., the elements of A ∩B are included twice. Thus to get the probability P(A∪B) we have to subtract the probabilities of the sample points in A ∩ B from P(A) + P(B) 
i.e. P(A∪ B) = P(A)+ P(B) -
` P(ω_i ), ∀ ω_i  ∈ A ∩ B `
= = P(A)+ P(B) - P( A ∩ B) 
Thus we observe that, P(A ∪ B)= P(A)+ P(B)- P(A ∩ B).
Alternatively, it can also be proved as follows: 
A ∪ B = A ∪ (B – A), where A and B – A are mutually exclusive, 
and   B = (A ∩ B) ∪ (B – A), where A ∩ B and B – A are mutually exclusive. 
Using  Axiom (iii) of probability, we get 
P (A ∪B) = P (A) + P (B – A) ... (2) and 
P(B) = P ( A ∩ B) + P (B – A) ... (3) 
Subtracting (3) from (2) gives 
P (A ∪ B) – P(B) = P(A) – P (A ∩ B) or 
P(A ∪ B) = P(A) + P (B) – P (A ∩ B) 
The above result can further be verified by observing the Venn Diagram Fig. 

If A and B are disjoint sets, i.e., they are mutually exclusive events, then A ∩ B = φ
Therefore P(A∩ B) = P ( φ) = 0
Thus, for mutually exclusive events A and B, we have 
P(A∪ B)= P(A) + P(B),
 which is Axiom(iii) of probability.

4)  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` 1/10`
Now P(A) = P(2) + P(4) + P(6) + P(8)
= `1/10 + 1/10 + 1/10 + 1/10  = 4/10 = 2/5`
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)
`= 6/10 = 3/5`
Thus P(A') = `3/5 = 1-2/5` = 1- P(A)