**Operation of events in probability :**

Applying the concept of set theory, we can give a new dimension of the classical definition of probability.

A sample space may be defined as a non-empty set containing all the elementary events of a random experiment as sample points.

A sample space is denoted by S.

An event A may be defined as a non-empty subset of S. This is shown in the figure given below.

As for example, if a die is rolled once than the sample space is given by S = {1, 2, 3, 4, 5, 6}.

Next, if we define the events A, B and C such that

A = {x : x is an even no. of points in S}

B = {x : x is an odd no. of points in S}

C = {x : x is a multiple of 3 points in S}

Then, it is quite obvious that

A = {2, 4, 6}, B = {1, 3, 5} and C = {3, 6}.

The classical definition of probability may be defined in the following way.

Let us consider a finite sample space S. That is, a sample space with a finite no. of sample points, n (S).

We assume that all these sample points are equally likely. If an event A which is a subset of S, contains n (A) sample points, then the probability of A is defined as the ratio of the number of sample points in A to the total number of sample points in S.

P(A) = n(A)/n(S)

Union of two events A and B is defined as a set of events containing all the sample points of event A or event B or both the events. This is shown in the figure given below. And we have

AnB = { x : x ∈ A on x ∈ B } where x denotes the sample points.

Showing the union of two events A and B and also their intersection

In the above example, we have

AUC = {2, 3, 4, 6} and AUB = {1, 2, 3, 4, 5, 6}.

The intersection of two events A and B may be defined as the set containing all the sample points that are common to both the events A and B.

This is shown in the above figure. we have

AnB = { x : x∈A and x∈B }

In the above example, AnB = Null set

AnC = { 6 }

Since the intersection of the events A and B is a null set, it is obvious that A and B are mutually exclusive events as they cannot occur simultaneously.

The difference of two events A and B, to be denoted by "A–B", may be defined as the set of sample points present in set A but not in B.

That is,

A - B = { x : x∊A and x∉B }

Similarly, B - A = { x : x∊B and x∉A }

In the above examples,

A - B = Null set and A - C = { 2, 4 }

This is shown in the figure given below.

Showing (A – B) and (B – A)

The complement of an event A may be defined as the difference between the sample space S and the event A.

A' = { x : x∊S and x∉A }

A' = S - A

A' = { 1, 3, 5 }

The figure given below depicts A'

Now we are in a position to redefine some of the terms we have already discussed.

Two events A and B are mutually exclusive if

P (AnB) = 0

Or more precisely,

P(AuB) = P(A) + P(B)

Similarly three events A, B and C are mutually exclusive if

P(AuBuC) = P(A) + P(B) + P(C)

Two events A and B are exhaustive if

P(AuB) = 1

Similarly three events A, B and C are exhaustive if

P(AuBuC) = 1

Three events A, B and C are equally likely if

P(A) = P(B) = P(C)

Three events A, B and C are mutually exclusive, exhaustive and equally likely.

What is the probably of the complementary event of A?

**Solution :**

Since A, B and C are mutually exclusive, we have

P(AuBuC) = P(A) + P(B) + P(C) ------ (1)

Since they are exhaustive,

P(AuBuC) = 1 ------ (2)

Since they are equally likely events,

P(A) = P(B) = P(C) = K, Say ------ (3)

Combining equations (1), (2) and (3), we have

1 = K + K + K

1 = 3K

1/3 = K

Thus P(A) = P(B) = P(C) = 1/3

Therefore, P(A') = 1 - P(A)

P(A') = 1 - 1/3

**P(A') = 2/3**

After having gone through the stuff given above, we hope that the students would have understood "Operation of events in probability".

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