Axiomatic approach to probability
In earlier topics, we have considered random experiments, sample space and events associated with these experiments. In our day to day life we use many words about the chances of occurrence of events. Probability theory attempts to quantify these chances of occurrence or non occurrence of events.
Axiomatic approach is another way of describing probability of an event. In this approach some axioms or rules are depicted to assign probabilities.
Let S be the 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
(i) For any event E, P (E) ≥ 0 (ii) P (S) = 1
(iii) If E and F are mutually exclusive events, then P(E ∪ F) = P(E) + P(F).
It follows from (iii) that P(φ) = 0. To prove this, we take F = φ and note that E and φ are disjoint events. Therefore, from axiom (iii), we get
P (E ∪ φ) = P (E) + P (φ) or P(E) = P(E) + P (φ) i.e. P (φ) = 0.
Let S be a sample space containing outcomes ω1 , ω2 ,…,ωn , i.e.,
S = {ω1, ω2, …, ωn}
It follows from the axiomatic definition of probability that
(i) 0 ≤ P (ωi) ≤ 1 for each ωi ∈ S
(ii) P (ω1) + P (ω2) + … + P (ωn) = 1
(iii) For any event A, P(A) = Σ P(ωi ), ωi ∈ A.
Note:- It may be noted that the singleton {ωi} is called elementary event and for notational convenience, we write P(ωi ) for P({ωi }).
For example, in ‘a coin tossing’ experiment we can assign the number to each of the outcomes H and T.
i.e P(H) = — and P(T) = — (1)
Clearly this assignment satisfies 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 = ½, and probability of T = ½
if we take P(H) = 1/4 and P(T) = 3/4 (2)
Does this assignment satisfy the conditions of axiomatic approach?
Yes, in this case, probability of H = – and probability of T = 3/4.
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. Hence, we can say that there are many ways (rather infinite) to assign probabilities to outcomes of an experiment.
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)’. We may get 0, 1, 2 or 3 defective pens as result of this examination.
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}
Let us consider another experiment of ‘tossing a coin “twice”
The sample space of this experiment is S = {HH, HT, TH, TT}
Let the following probabilities be assigned to the outcomes
P(HH)= 1/4, P(HT)= 1/7, P(TH) = 2/7, P(TT) = 9/28
Clearly this assignment satisfies the conditions of axiomatic approach. Now, let us find the probability of the event E: ‘Both the tosses yield the same result’.
Here E = {HH,TT}
= P(HH) + P(TT) = 1/4 + 9/28= 4/7
For the event F: ‘exactly two heads’, we have F = {HH}
and P(F) = P(HH) = –
Probabilities of equally likely outcomes
Let a sample space of an experiment be
S = {ω1, ω2,…, ωn}.
Let all the outcomes are equally likely to occur, i.e., the chance of occurrence of each simple event must be same.
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
Probability of the event ‘A or B’
Let us now 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
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)
In general, if A and B are any two events associated with a random experiment, then by the definition of probability of an event, we have
Hence 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 exclsuive.
Using Axiom (iii) of probability, we get
(2)
(3)
The above result can further be verified by observing the Venn Diagram (Fig 16.1)
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.
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
Also event ‘not A’ = A′ = {1, 3, 5, 7, 9, 10}
Also, we know that A′ and A are mutually exclusive and exhaustive events i.e.,
