Mutually exclusive events are those events that do not occur at the same time. For example, when a coin is tossed then the result will be either head or tail, but we cannot get both the results. Such events are also called disjoint events since they do not happen simultaneously. If A and B are mutually exclusive events then its probability is given by P(A Or B) or P (A U B). Let us learn the formula of P (A U B) along with rules and examples here in this article.

What are Mutually Exclusive Events?

In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. In other words, mutually exclusive events are called disjoint events. If two events are considered disjoint events, then the probability of both events occurring at the same time will be zero.

If A and B are the two events, then the probability of disjoint of event A and B is written by:

Probability of Disjoint (or) Mutually Exclusive Event = P ( A and B) = 0

How to Find Mutually Exclusive Events?

In probability, the specific addition rule is valid when two events are mutually exclusive. It states that the probability of either event occurring is the sum of probabilities of each event occurring. If A and B are said to be mutually exclusive events then the probability of an event A occurring or the probability of event B occurring that is P (a ∪ b) formula is given by P(A) + P(B), i.e.,

• P (A Or B) = P(A) + P(B)
• P (A ∪ B) = P(A) + P(B)

Note:

If the events A and B are not mutually exclusive, the probability of getting A or B that is P (A ∪ B) formula is given as follows:

P (A ∪ B) = P(A) + P(B) – P (A and B)

Real-life Examples on Mutually Exclusive Events

Some of the examples of the mutually exclusive events are:

• When tossing a coin, the event of getting head and tail are mutually exclusive. Because the probability of getting head and tail simultaneously is 0.
• In a six-sided die, the events “2” and “5” are mutually exclusive. We cannot get both the events 2 and 5 at the same time when we threw one die.
• In a deck of 52 cards, drawing a red card and drawing a club are mutually exclusive events because all the clubs are black.

Dependent and Independent Events

Two events are said to be dependent if the occurrence of one event changes the probability of another event. Two events are said to be independent events if the probability of one event does not affect the probability of another event. If two events are mutually exclusive, they are not independent. Also, independent events cannot be mutually exclusive.

Rules for Mutually Exclusive Events

In probability theory, two events are mutually exclusive or disjoint if they do not occur at the same time. A clear case is the set of results of a single coin toss, which can end in either heads or tails, but not for both. While tossing the coin, both outcomes are collectively exhaustive, which suggests that at least one of the consequences must happen, so these two possibilities collectively exhaust all the possibilities. Though, not all mutually exclusive events are commonly exhaustive. For example, the outcomes 1 and 4 of a six-sided die, when we throw it, are mutually exclusive (both 1 and 4 cannot come as result at the same time) but not collectively exhaustive (it can result in distinct outcomes such as 2,3,5,6).

From the definition of mutually exclusive events, certain rules for probability are concluded.

• Addition Rule: P (A + B) = 1
• Subtraction Rule: P (A U B)^’ = 0
• Multiplication Rule: P (A ∩ B) = 0

There are different varieties of events also. For instance, think of a coin that has a Head on both the sides of the coin or a Tail on both sides. It doesn’t matter how many times you flip it, it will always occur Head (for the first coin) and Tail (for the second coin). If we check the sample space of such experiment, it will be either { H } for the first coin and { T } for the second one. Such events have single point in the sample space and are called  “Simple Events”. Such kind of two sample events is always mutually exclusive.

Conditional Probability for Mutually Exclusive Events

Conditional probability is stated as the probability of an event A, given that another event B has occurred. Conditional Probability for two independent events B has given A is denoted by the expression P( B|A) and it is defined using the equation

P(B|A)= P (A ∩ B)/P(A)

Redefine the above equation using multiplication rule: P (A ∩ B) = 0

P(B|A)= 0/P(A)

So the conditional probability formula for mutually exclusive events is:

P (B | A) = 0

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Solved Examples

Here the sample problem for mutually exclusive events is given in detail. Go through once to learn easily.

Question 1: What is the probability of a die showing a number 3 or number 5?

Solution: Let,

P(3) is the probability of getting a number 3

P(5) is the probability of getting a number 5

P(3) = 1/6 and P(5) = 1/6

So,

P(3 or 5) = P(3) + P(5)

P(3 or 5) = (1/6) + (1/6) = 2/6

P(3 or 5) = 1/3

Therefore, the probability of a die showing 3 or 5 is 1/3.

Question 2: Three coins are tossed at the same time. We say A as the event of receiving at least 2 heads. Likewise, B denotes the event of getting no heads and C is the event of getting heads on the second coin. Which of these is mutually exclusive?

Solution:  Firstly, let us create a sample space for each event. For the event ‘A’ we have to get at least two head. Therefore, we have to include all the events that have two or more heads.

Or we can write:

A = {HHT, HTH, THH, HHH}.

This set A has 4 elements or events in it i.e. n(A) = 4

In the same way,  for event B, we can write the sample as:

B = {TTT} and n(B) = 1

Again using the same logic, we can write;

C = {THT, HHH, HHT, THH} and n(C) = 4

So B & C and A & B are mutually exclusive since they have nothing in their intersection.

Question 3: The likelihood of the 3 teams a, b, c winning a football match are 1 / 3, 1 / 5 and 1 / 9 respectively. Find the probability that

a] out of the three teams, either team a or team b will win

b] either team a or team b or team c will win

c] none of the teams will win the match

d] neither team a nor team b will win the match

Answer: 

a) P (A or B will win) = 1/3 + 1/5 = 8/15

b) P (A or B or C will win) = 1/3 + 1/5 + 1/9 = 29/45

c) P (none will win) = 1 – P (A or B or C will win) = 1 – 29/45 = 16/45

d) P (neither A nor B will win) = 1 – P(either A or B will win)

= 1 – 8/15

= 7/15

Question 4: If A and B are two independent events, then A and B’ is:

Answer:  A ∩ B’ and A ∩ B are mutually exclusive events such that;

A = (A ∩ B’) ∪ (A ∩ B)

P(A) = P(A ∩ B’) + P(A ∩ B)

P(A ∩ B’) = P(A) – P(A ∩ B)

= P(A) – P(A).P(B)   (Since A and B are independent)

= P(A ∩ B’)

=> P(A) (1 – P(B)) = P(A) P(B’)

Thus, A and B’ are also independent.

Question 5: If P (A) = 2 / 3, P (B) = 1 / 2 and P (A ∪ B) = 5 / 6 then events A and B are:

Answer: 

P (A ∪ B) = P (A) + P (B) − P (A ∩ B)

5 / 6 = (2 / 3) + (1 / 2) − P (A ∩ B)

⇒ P (A ∩ B) = 0

The events A and B are mutually exclusive.

Question 6: A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that the card drawn is a king or an ace.

Answer: 

As per the definition of mutually exclusive events, selecting an ace and selecting a king from a well-shuffled deck of 52 cards are termed mutually exclusive events.

Assume X to be the event of drawing a king and Y to be the event of drawing an ace.

In a standard deck of 52 cards, there exists 4 kings and 4 aces.

P (an event) = count of favourable outcomes / total count of outcomes

P (selecting a king from a standard deck of 52 cards) = P (X) = 4 / 52 = 1 / 13

P (selecting an ace from a standard deck of 52 cards) = P (Y) = 4 / 52 = 1 / 13

To compute P (king or ace).

By the formula of addition theorem for mutually exclusive events,

P (X U Y) = P (X) + P (Y)

P (X U Y) = (1 / 13) + (1 / 13)

= (1 + 1) / 13

= 2 / 13

The probability of selecting a king or an ace from a well-shuffled deck of 52 cards = 2 / 13.

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