HOW TO CHECK THE ASSIGNMENT OF PROBABILITY IS PERMISSIBLE

Here we are going to see some example problems to check the assignment of probability is permissible.

Before going to see examples, let us look into the definition of mutually exclusive and exhaustive events.

A₁, A₂, A₃,.........A_K are called mutually exclusive and exhaustive events if, 

(i)  A_i n A_j  ≠ ∅

(ii)  A₁ U A₂ U A₃ U......A_k  =  S

An experiment has the four possible mutually exclusive and exhaustive outcomes A, B, C, and D. Check whether the following assignments of probability are permissible.

Question 1 :

P(A)  =  0.15, P(B)  =  0.30, P(C)  =  0.43, P(D)  =  0.12

Solution :

Since the experiment has exactly the three possible mutually exclusive outcomes A, B, C, D they must be exhaustive events.

S  =  A U B U C U D

So, by axioms of probability

P(A) ≥ 0, P(B) ≥ 0, P(C) ≥ 0 and P(D) ≥ 0

P(A U B U C U D) = P(A) + P(B) + P(C) + P(D)  =  P(S)  =  1

  =   0.15 + 0.30 + 0.43 + 0.12

  =  1

Probability of an event can be 1.

So, the assignment of probability is permissible.

Question 2 :

P(A)  =  0.22, P(B)  =  0.38, P(C)  =  0.16, P (D)  =  0.34

Solution :

So, by axioms of probability

P(A) ≥ 0, P(B) ≥ 0, P(C) ≥ 0 and P(D) ≥ 0

P(A U B U C U D)  =  P(A) + P(B) + P(C) + P(D)  =  P(S)  =  1

  =   0.22 + 0.38 + 0.16 + 0.34

  =  1.1

Probability of an even can not be greater than 1. 

So, the assignment of probability is not permissible.

Question 3 :

P(A)  =  2/5, P(B)  =  3/5, P(C)  =  -1/5, P(D)  =  1/5

Solution :

So, by axioms of probability

P(A) ≥ 0, P(B) ≥ 0 but P(C) ≤ 0

Probability of an event can not be negative.

So, the assignment of probability is not permissible.

Question 4 :

P(A)  =  1/√3, P(B)  =  1 - (1/√3), P(C)  =  0

Solution :

So, by axioms of probability

P(A) ≥ 0, P(B) ≥ 0, and P(C) ≥ 0

P(A U B U C)  = P(A) + P(B) + P(C)  =  P(S)  =  1

  =   1/√3 + [1 - (1/√3)] + 0

  =  1

Probability of an event can be 1.

So, the assignment of probability is permissible.

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