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Axiomatic Approach To Probability (Part I)

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Axiomatic Approach To Probability (Part I) - Lesson Summary

An event is an outcome or a collection of outcomes.

In the classical theory,

Probability of an event =  Number of outcomes favourable to the event Total number of possible outcomes

In this theory, the number of outcomes is finite and all the outcomes are equally likely.

In the statistical theory of probability, we find the probability of an event happening based on the observations and the data collected. Neither theory can be applied to experiments that have an infinite number of outcomes.

All outcomes are not necessarily equally likely.

Example, the event of getting at least a head in an experiment of tossing of two coins simultaneously.

A. N. Kolmogorov, a Russian mathematician, developed a new theory of probability in 1933 called the axiomatic approach to probability.

In the axiomatic approach to probability, the probability is defined by some axioms or rules.

Let S be the sample space of a random experiment.

In the axiomatic approach to probability, probability P is a real valued function, whose domain is the set of all the subsets of sample space S, and the range is [0, 1].

If the domain is defined by P(S) and the range is [0, 1], then P:P(S) → [0,1 ].

P satisfies the following axioms:

The probability of any event, say A, is non-negative i.e. P(A) ≥ 0

The probability of sample space S is 1 i.e. P(S) = 1

If A and B are mutually exclusive events, then P (A ∪ B) = P (A) + P (B).

Consider an experiment, where a die is rolled.

The sample space associated with this experiment is S = {1, 2, 3, 4, 5, 6}.

P({1}) = 1 6

P({2}) = 1 6

P({3}) =  1 6

P({4}) = 1 6

P({5}) = 1 6

P({6}) = 1 6

The probabilities of the events are equal, which is equal to 1/6. The probability of each event lies between zero and one.
Hence, the probabilities of the events satisfy axiom one.

Sum of the probabilities of all the events is equal to 1.

P({1)} + P({2)} + P({3)} + P({4)} + P({5)} + P({6)} =   1 6    +  1 6    +  1 6    +  1 6    +  1 6    +  1 6    = 1 .

This implies that the probability of S is equal to one.
P({1, 2, 3, 4, 5, 6)} = P(S) = 1.

Hence, the probabilities of the sample space satisfy axiom two.

Consider that a coin is tossed once. The chance of getting a head is trebled as compared to the chance of getting a tail when a coin is tossed twice.

P(H) = 3 4  

P(T) =   1 4   

P(H) =   3 4      0 ≤   3 4    ≤ 1

P(T) =  1 4      0 ≤   1 4    ≤ 1

P(H) + P(T) = 3 4   +   1 4       = 1

P(S) = 1

Theorems related to probability:

Theorem: The probability of an event that cannot occur (impossible event) is zero, i.e. P (f) = 0.

Proof:

Consider two events A and B, such that A = B =  ∅.

Thus, A ∩ B = ∅ ∩ ∅ =  ∅

⇒ Events A and B are mutually exclusive.

Hence, P(A ∪ B) = P(A) + P(B)

⇒ P(∅ ∪ ∅) = P(∅) + P(∅)

⇒ P(∅) = P(∅) + P(∅)

⇒ P(∅) = 0

Hence, the probability of an event that cannot occur is zero.

Theorem: If A is an event in sample space S, then 0 ≤ P (A) ≤ 1.

Proof:

A is an event in S.

∅ ⊆ A ⊆ S

P(f) ≤ P(A) ≤ P(S)

⇒ 0 ≤ P(A) ≤ 1.

Hence, the probability of an event of a sample space always lies between zero and one.

Theorem: If E is an event in sample space S containing n sample points that are equally likely, then P (E) = 1n .

Proof:

Let P(E) = x

E1, E2, E3,..., En are the elementary events that contain each sample point in sample space S.

Thus, the events are mutually exclusive and exhaustive.

∴ E1 ∪ E2 ∪ E3 ∪...∪ En = S

P(E1 ∪ E2 ∪ E3 ∪...∪ En) = P(S)

⇒ P(E1) + P(E2) + P(E3) + ... P(En) = P(S)

Sample space S contains equally likely sample points.

⇒ P(E1) = P(E2) = P(E3) = ... = P(En) = x

⇒ x + x + x + .....+ x (n times) = 1

⇒ nx = 1

⇒ x = 1n

Hence, P(E) = 1n.

Theorem: S is a sample space containing n sample points and A is an event in S containing m sample points. If each outcome is equally likely, then P(A) = mn.

Proof:

Let E1, E2, E3,...Em be the elementary events of S containing each of the m sample points of A.

These events are mutually exclusive and exhaustive.

∴ E1 ∪ E2 ∪ E3 ∪...∪ Em = A

P(E1 ∪ E2 ∪ E3 ∪...∪ Em) = P(A)

⇒ P(A) = P(E1) + P(E2) + P(E3) + ... P(Em)

=  1n + 1n + 1n + .... + 1n (m times)    (∵ S contains n sample points.)

= mn

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