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Statistics

Mutually Exclusive Events

Two sets are known to be mutually exclusive when they have no common elements. 

Consider the set of all even positive integers, and the set of all odd positive integers: 

Set A = {\( 2, 4, 6, 8, 10, 12, 14, 16... \)}

Set B = {\( 1, 3, 5, 7, 9, 11, 13, 15... \)}

We call them mutually exclusive since none of the elements of Set A are in Set B, and vice versa. 

Recall that an event is a set of outcomes from Simple and Compound Events. It follows that mutually exclusive events are those that do not share any of the same outcomes. 

How do we calculate the probability of these events? Let us visualize using a Venn Diagram:

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If \( A \) and \( B \) are two mutually exclusive events, then the probability of \(A \) or \( B \) occurring is their respective probabilities added together. 

Non-Mutually Exclusive Events

Two sets are non-mutually exclusive if they share common elements. 

Consider the set of all numbers from 1 to 10, and the set of all even numbers from 1 to 16:

Set A = {\( 2, 4, 6, 8, 10, 12, 14, 16 \)}

Set B = {\( 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \)}

We call them non-mutually exclusive since they share the common elements of \( 2, 4, 6 \) and \( 8 \). 

It follows that two events are non-mutually exclusive if they share common outcomes. 

How do we calculate the probability of these events? Let us visualize using another Venn Diagram:

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If \( A \) and \( B \) are two non-mutually exclusive events, then the probability of \( A \) or \(B \) occuring is both of their probabilities added together and subtracting the probability of both of them occurring. 

EXAMPLE 

a) A box contains 2 red, 4 green, 5 blue and 3 yellow marbles. If a single random marble is chosen from the box, what is the probability that it is red or green marble?

b) In a math class of 30 students, 17 are boys and 13 are girls. On a unit test, 4 boys and 5 girls made an A grade. If a student is chosen at random from the class, what is the probability of choosing a girl or an A student?

See the video below for the solution:

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