In normal approach to probability, we consider random experiments, sample space and other events that are associated with the different experiments. In our day to day life, we are more familiar with the word ‘chance’ as compared to the word ‘probability’. Since, Mathematics is all about quantifying things, the theory of probability basically quantifies these chances of occurrence or non-occurrence of the events.

One important thing about probability is that it can only be applied to experiments where we know the total number of outcomes of the experiment, i.e. unless and until we know the total number of outcomes of an experiment, concept of probability cannot be applied.

Thus, in order to apply probability in day to day situations, we should know total number of possible outcomes of the experiment. **Axiomatic Probability** is just another way of describing the probability of an event. As, the word itself says, in this approach some axioms are predefined before assigning probabilities. This is done to quantize the event and hence to ease the calculation of occurrence or non-occurrence of the event.

Let, S be the sample space of any random experiment and let \(P\) be the probability of occurrence of any event. Noting the characteristics of \(P\), it should be a real valued function whose domain will be the power set of \(S\) and the range will lie in the interval \([0,1]\). This probability \(P\) will satisfy the following probability axioms:

1. For any event \(E\), \(P(E) ≥ 0\)

2. \(P(S)\) = \(1\)

3. In case \(E\) and \(F\) are mutually exclusive events, then following equation will be valid:

\(P(E ∪ F)\) = \(P(E) + P(F)\)

From point (3) it can be stated that \(P(ф)\) = \(0\)

If we need to prove this, let us take \(F\) = \(ф\) and make a note that \(E\) and \(ф\) are disjoint events. Hence, from point (3) we can deduce that

\(P(E ∪ ф)\) = \(P(E) + P(ф)\) or \(P(E)\) = \(P(E) + P(ф)\) i.e. \(P(ф)\) = \(0\)

Let, the sample space of S contains the given outcomes \(δ_1, δ_2, δ_3 …… δ_n\), then as per axiomatic definition of probability, we can deduce the following points:

- \(0 ≤ P(δ_i) ≤ 1\) for each \(δ_i ∈ S\)
- \(P(δ_1) + P(δ_2)+ … + P(δ_n)\) = \(1\)
- For any event \(Q\), \(P(Q)\) = \(∑P(δ_i), δ_i ∈ Q\).

This is to be noted that the singleton {\(δ_i\)} is known as **elementary event** and for the convenience of writing, we write \(P(δ_i)\) for \(P\)({\(δ_i\)}) .

**Example**

On tossing a coin we say that the probability of occurrence of head and tail is \(\frac{1}{2}\) each. Basically here we are assigning the probability value of \(\frac{1}{2}\) for the occurrence of each event.

This condition satisfies both the conditions, i.e.

- Each value is neither less than zero nor greater than 1
- Sum of the probabilities of occurrence of head and tail is 1

Hence, for this case we can say that the probabilities of occurrence of head and tail are \(\frac{1}{2}\) each.

Now, say \(P(H)\) = \(\frac{5}{8}\) and \(P(T)\) = \(\frac{3}{8}\)

Does this probability value satisfy the conditions of axiomatic approach?

For this, let us again check the basic initial conditions of axiomatic approach of probability.

- Each value is neither less than zero nor greater than 1
- Sum of the probabilities of occurrence of head and tail is 1

Hence, this sort of probability value assignment also satisfies the axiomatic approach of probability. Thus, we can conclude that there can be infinite ways to assign probability to outcomes of an experiment.