Lesson Explainer: Mutually Exclusive Events Mathematics

In this explainer, we will learn how to identify mutually exclusive events and nonmutually exclusive events and find their probabilities.

You may recall that if events are mutually exclusive or disjoint, they cannot occur at the same time. For example, an animal cannot be both a cat and a dog: “being a cat” and “being a dog” are mutually exclusive or disjoint events. However, a person may like both cats and dogs, so “likes cats” and “likes dogs” are not mutually exclusive or disjoint events.

To calculate probabilities for mutually exclusive and nonmutually exclusive events, we will need to use some of the rules of probability. Let us remind ourselves of these.

Some Probability Rules and Definitions

For any event 𝐴, if 𝑃(𝐴) is the probability of event 𝐴 occurring, we have the following:

Rule 1

0𝑃(𝐴)1

Rule 2

The sum of the probabilities of all possible outcomes is equal to 1 (or 100%).

The complement of event 𝐴, written as 𝐴, refers to everything that is not  𝐴.

Rule 3

𝑃(𝐴)=1𝑃(𝐴)

Note: The complement of event 𝐴 is sometimes also written as 𝐴. Also note that 𝑃(𝐴𝐵) refers to the probability of the occurrence of 𝐴 and the nonoccurrence of 𝐵.

If two events 𝐴 and 𝐵 cannot occur at the same time, we say that they are mutually exclusive or disjoint events. In this case, the joint probability of 𝐴 and 𝐵 occurring at the same time is zero. We write this as 𝑃(𝐴𝐵)=0. If events 𝐴 and 𝐵 are mutually exclusive, the probability that 𝐴 or 𝐵 occurs is the sum of their probabilities.

That is,

Rule 4

𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)

If events 𝐴 and 𝐵 are not mutually exclusive, the probability that either 𝐴 or 𝐵 or both occur is

Rule 5

𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵).

We are now armed with all the tools we need to tackle some examples. This first example uses the “total probability” rule for mutually exclusive events.

Example 1: Total Probability and Mutually Exclusive Events

In an animal rescue shelter, 39% of the current inhabitants are cats (C) and 41% are dogs (D).

  1. Find the probability that an animal chosen at random is either a cat or a dog.
  2. Find the probability that an animal chosen at random is neither a cat nor a dog.

Answer

Part 1

To find the probability that an animal chosen at random is either a cat or a dog, we note that the events “cat” and “dog” are mutually exclusive. That is, an animal can be either a cat or a dog but not both. This means that 𝑃()=0CD. We can illustrate this in a Venn diagram as below.

If we let 𝑃()C be the probability that an animal chosen at random is a cat and let 𝑃()D be the probability that an animal chosen at random is a dog, converting our percentages into probabilities by dividing by 100, we have 𝑃()=0.39C and 𝑃()=0.41D. We can now use the probability rule which states that, for mutually exclusive events, 𝑃()=𝑃()+𝑃()CDCD This gives us 𝑃()=𝑃()+𝑃()=0.39+0.41=0.8.CDCD

Hence, the probability that an animal chosen at random is either a cat or a dog is 0.8. Multiplying by 100 to get back to a percentage, we have an 80% chance of selecting either a cat or a dog.

Part 2

To find the probability that an animal chosen at random is neither a cat nor a dog, we use the rule that the sum of the probabilities of all possible outcomes is equal to 1. We know that the probability of either a cat or a dog is 0.8, so the probability of selecting neither a cat nor a dog is 10.8=0.2. As a percentage (×100), this means that 20% must be neither cats nor dogs.

The next example tests our understanding of mutual exclusivity.

Example 2: Are the Events Mutually Exclusive?

Farida has a deck of 52 cards.

She randomly selects one card and considers the following events:

  • Event 𝐴: Picking a card that is a heart;
  • Event 𝐵: Picking a card that is black;
  • Event 𝐶: Picking a card that is not a spade.

  1. Are events 𝐴 and 𝐵 mutually exclusive?
  2. Are events 𝐴 and 𝐶 mutually exclusive?
  3. Are events 𝐵 and 𝐶 mutually exclusive?

Answer

Part 1

To determine whether event 𝐴, “picking a card that is a heart,” and event 𝐵, “picking a card that is black,” are mutually exclusive or not, let us mark the two events on a diagram.

Event 𝐴 is enclosed in the red rectangle and event 𝐵 is enclosed in the green rectangle. There are no cards enclosed by both the red and green rectangles. This means that no card can be both a heart and a black card, so the two events are mutually exclusive.

Part 2

To determine whether event 𝐴, “picking a card that is a heart,” and event 𝐶,“picking a card that is not a spade,” are mutually exclusive, we will leave our red rectangle around the heart cards and shade all the cards that are not spades blue.

We can see that the cards within the red rectangle are also shaded blue, so a card can be a heart and not a spade at the same time. Hence, in this case, the events are not mutually exclusive.

Part 3

To determine whether event 𝐵, “picking a card that is black,” and event 𝐶, “picking a card that is not a spade,” are mutually exclusive or not, we will keep our blue shading for cards that are not spades (𝐶) and enclose cards that are black (𝐵) within a purple rectangle.

We can see from our diagram that there are cards both shaded blue and enclosed within the purple rectangle. So a card can be black and not a spade. (All the “club” cards belong to the intersection of the two events.) Hence, events 𝐵 and 𝐶 are not mutually exclusive.

In our next example, we will calculate the probability for mutually exclusive events.

Example 3: Calculating Probability for Mutually Exclusive Events

A small choir has a tenor singer, 3 soprano singers, a baritone singer, and a mezzo-soprano singer. If one of their names was randomly chosen, determine the probability that it was the name of the tenor singer or a soprano singer.

Answer

Our choir consists of the singers shown.

If we assume that the singers stick to their parts so that, for example, a soprano singer does not sing tenor or baritone parts and vice versa, then the events “soprano,” “tenor,” “baritone,” and “mezzo-soprano” are mutually exclusive.

This being the case, to find the probability that a randomly chosen singer is either a tenor or a soprano singer, we can use the probability rule: 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵), since, by mutual exclusion, 𝑃(𝐴𝐵)=0.

As there are 6 singers in total and only one is a tenor, the probability that a randomly chosen singer is a tenor is 𝑃()==16.Tenornumberoftenorstotalnumberofsingers

Similarly, for the soprano singers, there are 3 sopranos; hence, 𝑃()==36.Sopranonumberofsopranostotalnumberofsingers

Applying the rule that, for mutually exclusive events, 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵), we have 𝑃()=𝑃()+𝑃()=16+36=46=23.TenorSopranoTenorSoprano

The probability that a singer chosen at random is either a tenor or a soprano is therefore 23=0.667, to 3 decimal places. We can say then that there is a 67% chance that a singer chosen at random is either a tenor or a soprano (0.667×100=66.7%).

In our next example, we will calculate the probability for mutually exclusive events.

Example 4: Probability for Mutually Exclusive Events

Suppose that 𝐴 and 𝐵 are two mutually exclusive events. The probability of event 𝐵 occurring is five times that of event 𝐴 occurring. Given that the probability that one of the two events occurs is 0.18, find the probability of event 𝐴 occurring.

Answer

To find the probability of event 𝐴 occurring, we know that the probability of either 𝐴 or 𝐵 occurring (𝑃(𝐴𝐵)) is 0.18. And we know that 𝐴 and 𝐵 are mutually exclusive, so 𝑃(𝐴𝐵)=0. We also know that the probability of 𝐵 occurring is five times that of 𝐴 occurring, so 𝑃(𝐵)=5𝑃(𝐴) Putting all of this information into rule 4 for mutually exclusive events, we have 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)0.18=𝑃(𝐴)+5𝑃(𝐴)0.18=6𝑃(𝐴)0.186=𝑃(𝐴)0.03=𝑃(𝐴).i.e.,

Hence, 𝑃(𝐴)=0.03.

Our next example gives us more practice at working out probabilities for mutually exclusive events.

Example 5: Mutually Exclusive Events and Probability

Suppose 𝐴 and 𝐵 are two mutually exclusive events. Given that 𝑃(𝐴𝐵)=0.93 and 𝑃(𝐴𝐵)=0.39, find 𝑃(𝐵).

Answer

To find the probability of 𝐵, we note that since 𝐴 and 𝐵 are mutually exclusive events, they can never occur together. Hence, 𝑃(𝐴𝐵)=0.39 must equal 𝑃(𝐴).

Again, because 𝐴 and 𝐵 are mutually exclusive and cannot occur together, by rule 4, 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵).

We are given that this is 0.93. If we substitute this and our value for 𝑃(𝐴) into the equation above, we find 0.93=0.39+𝑃(𝐵).

Rearranging this gives us 𝑃(𝐵)=0.930.39=0.54. Hence, 𝑃(𝐵)=0.54.

In our next example, we will consider probabilities for nonmutually exclusive events.

Example 6: Probability and Nonmutually Exclusive Events

The probability that a student passes their physics exam is 0.71. The probability that they pass their mathematics exam is 0.81. The probability that they pass both exams is 0.68. What is the probability that the student only passes their mathematics exam?

Answer

To find the probability that a student passes mathematics (M) but not physics (P), let us illustrate the events in a Venn diagram.

We note that since there is overlap, the events are not mutually exclusive and can occur together. Now if we highlight the probabilities that we know concerning the event “passes math” on the diagram, that is, 𝑃()=0.81Math and 𝑃()=0.68MathPhysics, we have

The event “passes math” is everything inside the red oval, which has a probability of 0.81. The overlap in the center of the diagram covers “passes both math and physics” and has a probability of 0.68. But we want to find the probability of passing mathematics but not physics, which covers the dark purple section in the diagram below and is denoted 𝑃()MP.

If we put our information in an equation, we have 𝑃()=𝑃()+𝑃()0.81=𝑃()+0.68.MMPMPi.e.,MP

And rearranging this gives us 𝑃()=0.810.68=0.13.MP

Hence, the probability that a student passes mathematics but not physics is 0.13.

Our final example gives us more practice with nonmutually exclusive events.

Example 7: Probability for Nonmutually Exclusive Events

A survey asked 49 people if they had visited any clubs recently. 28 had attended club 𝐴, 38 had attended club 𝐵, and 8 had not been to either club. What is the probability that a random person from the sample attended both clubs?

Answer

To find the probability that a person chosen at random attended both clubs, let us collect the information we have:

  • There were 49 people in total.
  • 28 attended club 𝐴.
  • 38 attended club 𝐵.
  • 8 attended neither club 𝐴 nor club 𝐵.

Since there were 49 people and 8 attended neither club, we know then that 498=41 people attended at least one club.

We can use this information in probability rule 5: 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵), that is, the probability that a person chosen at random attended either club 𝐴 or club 𝐵 or both, to calculate the probability that a person chosen at random attended both clubs: 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵)4149=2849+3849𝑃(𝐴𝐵).

Rearranging this gives us 𝑃(𝐴𝐵)=2849+38494149=2549.

Hence, the probability that a person chosen at random attended both clubs is 25490.510, or approximately 50% of the people in the sample attended both clubs.

Let us finish our examination of the probability of mutually and nonmutually exclusive events with a reminder of the main points and rules we need.

Key Points

  • If two events 𝐴 and 𝐵 cannot occur at the same time, we say that they are mutually exclusive or disjoint events. In this case, the joint probability of 𝐴 and 𝐵 occurring at the same time is zero. We write this as 𝑃(𝐴𝐵)=0. If events 𝐴 and 𝐵 are mutually exclusive, the probability that 𝐴 or 𝐵 occurs is the sum of their probabilities.
    That is, 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵).
  • If events 𝐴 and 𝐵 are not mutually exclusive, the probability that either 𝐴 or 𝐵 or both occur is 𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵).

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