Let A, B and C be any two events associated with a random experiment whose sample space is S. Then, 

(i) A ∪ B. (Union of A and B) is the event that occurs if A occurs or B occurs or both A and B occur. 

(ii) A ∩ B. (Intersection of A and B). It is the event set which contains all sample points or outcomes which the two events A and B have in common.

Ex: (In a throw of a die), 

A : Event of getting an odd number 

B : Event of getting a prime number 

⇒ A = {1, 3, 5}, B = {2, 3, 5}. Then, A ∩ B = {3, 5}. 

(iii) \(\bar{A}\), (Complement of an event A). It is the set of all sample points of sample spaces that are not contained in A. 

In the toss of a coin, if A is getting a tail, then \(\bar{A}\) is getting a head. 

(iv) (A ∪ B ∪ C) is the event that occurs when at least one of the events A, B or C occurs 

(v) (A ∩ B ∩ C) is the event that occurs when all the three events A, B and C occur. 

(vi) As mutually exclusive events cannot occur together, if events A and B are mutually exclusive, then A ∩ B = ϕ since A and B have nothing in common. 

(vii) Mutually exclusive and exhaustive events. 

Let S be the sample space associated with a random experiment. If E₁, E₂, ..., En are mutually exclusive elementary events associated with the random experiment, then E_i ∩ E_j = ϕ for all i ≠ j and E₁ ∪ E₂ ∪ E₃... ∪ E_n = S 

(Since exhaustive means the total number of possible outcomes) Therefore, an event and its complementary event are both mutually exclusive and exhaustive since: 

A ∩ \(\bar{A}\) = ϕ and A ∪ \(\bar{A}\) = S. 

Ex. Let 1 ball be drawn from a bag containing 12 balls of which 4 balls are white, 4 are red and 4 are green. 

Let A : Event-ball drawn is white 

B : Event-ball drawn is red 

C : Event-ball drawn is green. 

It is obvious that one of the three events must occur as the ball drawn is either white or red or green. 

This means that A, B and C form a mutually exclusive and exhaustive set of events. 

∴ A∩B = ϕ, B∩C = ϕ, A ∩C = ϕ and A∪B∪C = S.