II. Theorem of Addition

Theorem of Addition is used when one has to determine the probability of occurrence of two or more events. In simple terms, this theorem is used to calculate the probability of union of two or more than two events. For instance, there are two events E₁ and E₂ of a given sample space. By using the theorem of addition, we can determine the probability that either E₁ or E₂ will occur. However, to determine the probability, first of all, we have to find out whether the events are mutually exclusive or overlapping, after that only the required probability is calculated using the correct rule or formula.

Theorem of Addition has two cases:

1. When the events are Mutually Exclusive

If A and B are mutually exclusive events then according to this theorem:

P(E₁ ⋃ E₂) = P(E₁) + P(E₂)

where, (E₁ ⋃ E₂) means either E₁ or E₂

If there are more than two events, then

P(E₁ ⋃ E₂ ⋃ …….. E_n) = P(E₁) + P(E₂) + ……… + P(E_n)

If these events are collectively exhaustive, then

P(E₁ ⋃ E₂ ⋃ …….. E_n) = P(E₁) + P(E₂) + ……… + P(E_n) = 1

This rule is known as the Theorem of Addition for Mutually Exclusive Events.

Example:

If A and B are events of occurrence of 2 and occurrence of 3 on a dice respectively, then calculate the probability of occurrence of A or B; i.e., getting 2 or 3 on dice.

Solution:

As A and B are independent and mutually exclusive events, using the theorem of addition,

P(A⋃B) = P(A) + P(B)

P(A⋃B)=

P(A⋃B)=

2. When the Events are Overlapping

When the events are overlapping, the theorem of addition determines the probability that one or more events would occur in a single trial.

If E₁ and E₂ are overlapping events then according to this theorem:

P(E₁ ⋃ E₂) = P(E₁) + P(E₂) – P(E₁ ⋂ E₂)

where, P(E₁ ⋃ E₂) is the probability that either E₁ or E₂ or both events will occur, and P(E₁ ⋂ E₂) is the joint probability that indicates the probability of occurrence of both E₁ and E₂.

If there are three events E₁, E₂, and E₃, then

P(E₁ ⋃ E₂ ⋃ E₃) = P(E₁) + P(E₂) + P(E₃) – P(E₁ ⋂ E₂) – P(E₁ ⋂ E₃) – P(E₂ ⋂ E₃) + P(E₁ ⋂ E₂ ⋂ E₃)

where P(E₁ ⋃ E₂ ⋃ E₃) gives the probability of occurrence of at least one of the events; E₁, E₂, and E₃.

Example:

Consider a class with 20 students. 10 students passed in Maths, 15 in English, and 13 in both. What is the probability that a student passed in either Math or English?

Solution:

Total students = 20

P(A) = Probability that a student passed in Maths = 

P(B) = Probability that a student passed in English = 

P(A⋂B) = Probability that a student passed in both Math and English = 

This is a case of overlapping events as some students who passed in Math may have passed in English too and some students who passed in English may have passed in Math too. Thus we need to remove these common students from the sum of students who passed in Math and English. Thus, using the theorem of addition for overlapping events, we get:

The probability that a student passed either in Maths or in English = P(A⋃B)

P(A⋃B) = P(A) + P(B) – P(A⋂B)

P(A⋃B) =

P(A⋃B) =

P(A⋃B) =