In probability two events are said to be mutually exclusive if and only if the events have no shared outcomes. If we consider the events as sets, then we would say that two events are mutually exclusive when their intersection is the empty set. We could denote that events A and B are mutually exclusive by the formula A ∩ B = Ø. As with many concepts from probability, some examples will help to make sense of this definition.
Suppose that we roll two six-sided dice and add the number of dots showing on top of the dice. The event consisting of "the sum is even" is mutually exclusive from the event "the sum is odd." The reason for this is because there is no way possible for a number to be even and odd.
Now we will conduct the same probability experiment of rolling two dice and adding the numbers shown together. This time we will consider the event consisting of having an odd sum and the event consisting of having a sum greater than nine. These two events are not mutually exclusive.
The reason why is evident when we examine the outcomes of the events. The first event has outcomes of 3, 5, 7, 9 and 11. The second event has outcomes of 10, 11 and 12. Since 11 is in both of these, the events are not mutually exclusive.
We illustrate further with another example. Suppose we draw a card from a standard deck of 52 cards. Drawing a heart is not mutually exclusive to the event of drawing a king. This is because there is a card (the king of hearts) that shows up in both of these events.
There are times when it is very important to determine if two events are mutually exclusive or not. Knowing whether two events are mutually exclusive influences the calculation of the probability that one or the other occurs.
Go back to the card example. If we draw one card from a standard 52 card deck, what is the probability that we have drawn a heart or a king?
First, break this into individual events. To find the probability that we have drawn a heart, we first count the number of hearts in the deck as 13 and then divide by the total number of cards. This means that the probability of a heart is 13/52.
To find the probability that we have drawn a king we start by counting the total number of kings, resulting in four, and next divide by the total number of cards, which is 52. The probability that we have drawn a king is 4/52.
The problem is now to find the probability of drawing either a king or a heart. Here’s where we must be careful. It is very tempting to simply add the probabilities of 13/52 and 4/52 together. This would not be correct because the two events are not mutually exclusive. The king of hearts has been counted twice in these probabilities. To counteract the double counting, we must subtract the probability of drawing a king and a heart, which is 1/52. Therefore the probability that we have drawn either a king or a heart is 16/52.
A formula known as the addition rule gives an alternate way to solve a problem such as the one above. The addition rule actually refers to a couple of formulas that are closely related to one another. We must know if our events are mutually exclusive in order to know which addition formula is appropriate to use.
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:
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:
Page 2
How many ways can we arrange the letters in the word MATH ?
Imagine we have four blank spaces that represent where each letter in the word MATH can go.
_ _ _ _
From here we can ask ourselves, how many choices of the letters in the word MATH can go in the first position?
Since there are 4 letters, there are 4 options; the arrangement can begin with M, A, T, or H.
Now let's imagine we chose one of those letters and locked it into the first position.
X _ _ _
or we can place a 4 here to represent the four options:
4 _ _ _
Now if we look to the next blank, how many options do we now have for the second position?
Since we already used one letter for the first blank, we are now left with 3 letters we can choose from.
X X _ _
or 4 3 _ _
If we continue and lock in one of the three letters into the second position, that leaves us with 2 options for the third position and subsequently one letter for the fourth position.
4 3 2 1
So how can we use this to figure out the number of unique arrangements?
Well, you can imagine this as an application of the product rule. We are choosing one letter for the first position AND one letter for the second position AND one letter for the third AND one letter for the fourth. Since we figured out how many options we have for each of those choices, we can use the product rule to find the number of arrangements.
\(4 \times 3 \times 2 \times 1 = 24\)
Then there are 24 arrangements. We call these arrangements permutations which we will talk more about in the next lesson.
This product where each term is decreasing by 1 until we reach 1 has a special name; we call this a factorial.
We represent a factorial with an exclamation mark like this:
\(4 \times 3 \times 2 \times 1 = 4!\)
Some things to note:
\(1! = 1\)
\( (n+1) \times n! = (n+1)! \)
For example:
\(5 \times 4! = 5!\)
or
\(90 \times 8! = 10 \times 9 \times 8! = 10!\)
Probability and Statistics > Probability > Mutually Exclusive Events
What is a Mutually Exclusive Event?
Mutually exclusive events are things that can’t happen at the same time. For example, you can’t run backwards and forwards at the same time. The events “running forward” and “running backwards” are mutually exclusive. Tossing a coin can also give you this type of event. You can’t toss a coin and get both a heads and tails. So “tossing a heads” and “tossing a tails” are mutually exclusive. Some more examples are: your ability to pay your rent if you don’t get paid, or watching TV if you don’t have a TV.
A coin toss can be mutually exclusive.
Watch the video for the definition and two examples of finding probabilities for mutually exclusive events:
Mutually Exclusive Events
Watch this video on YouTube.
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Mutually Exclusive Event Probability
The basic probability(P) of an event happening (forgetting mutual exclusivity for a moment) is:
P = Number of ways the event can happen / total number of outcomes.
Example: The probability of rolling a “5” when you throw a die is 1/6 because there is one “5” on a die and six possible outcomes. If we call the probability of rolling a 5 “Event A”, then the equation is:
P(A) = Number of ways the event can happen / total number of outcomes
P(A) = 1 / 6.
It’s impossible to roll a 5 and a 6 together; the events are mutually exclusive.
The events are written like this:
P(A and B) = 0
In English, all that means the probability of event A (rolling a 5) and event B (rolling a 6) happening together is 0.
However, when you roll a die, you can roll a 5 OR a 6 (the odds are 1 out of 6 for each event) and the sum of either event happening is the sum of both probabilities. In probability, it’s written like this:
P(A or B) = P(A) + P(B)
P(rolling a 5 or rolling a 6) = P(rolling a 5) + P(rolling a 6)
P(rolling a 5 or rolling a 6) = 1/6 + 1/6 = 2/6 = 1/3.
It’s impossible to roll a 1 and a 2 together.
Mutually Exclusive Event PRobability: Steps
Example problem: “If P(A) = 0.20, P(B) = 0.35 and (P A∪ B) = 0.51, are A and B mutually exclusive?”
Note: a union (∪) of two events occurring means that A or B occurs.
Step 1: Add up the probabilities of the separate events (A and B). In the above example:
.20 + .35 = .55
Step 2: Compare your answer to the given “union” statement (A ∪ B). If they are the same, the events are mutually exclusive. If they are different, they are not mutually exclusive. Why? If they are mutually exclusive (they can’t occur together), then the (∪)nion of the two events must be the sum of both, i.e. 0.20 + 0.35 = 0.55.
In our example, 0.55 does not equal 0.51, so the events are not mutually exclusive.
Like the explanation? Check out more step by step examples—just like this one for mutually exclusive events—in the Practically Cheating Statistics Handbook
References
Beyer, W. H. CRC Standard Mathematical Tables, 31st ed. Boca Raton, FL: CRC Press, pp. 536 and 571, 2002.
Dodge, Y. (2008). The Concise Encyclopedia of Statistics. Springer.
Everitt, B. S.; Skrondal, A. (2010), The Cambridge Dictionary of Statistics, Cambridge University Press.
Wheelan, C. (2014). Naked Statistics. W. W. Norton & Company
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