What if you rolled a pair of dice? How could you find the probability that you will roll either a 1 or a 12? After completing this Concept, you'll be able to find the probability of mutually exclusive events like this one.
Watch This
CK-12 Foundation: Mutually Exclusive Events
Guidance
Imagine you are going to see a movie. Your friend has just bought the tickets, and you are not sure which movie you are seeing. There are 4 movies playing. Harry Potter (which you have already seen, but your friend has not) is one of them.
- What are the chances you will be seeing Harry Potter?
- What are the chances you will NOT be seeing Harry Potter?
This is an easy example of a mutually exclusive event: you will either see Harry Potter, or you will not. You cannot do both!
Finding the probability of mutually exclusive events is easy; what’s not as easy is finding the probability of events that can overlap depend on each other. In this lesson, you’ll learn how to find the probability of any two events that can be related to each other.
Find the Probability of Mutually Exclusive Events
In probability, when two events are mutually exclusive, the probability of both happening together is zero.
Examples of mutually exclusive events in probability include:
- Flipping a coin and:
- getting heads
- getting tails
- Picking a single card from a deck and:
- getting an ace
- getting a 7
- getting a queen
- etc...
- Picking a single colored marble from a bag and:
- getting a red marble
- getting a blue marble
- getting a green marble
- etc..
What this means mathematically is two-fold. If our two mutually exclusive events are \begin{align*}A\end{align*} and \begin{align*}B\end{align*}:
- \begin{align*}P(A \ and \ B) = 0.\end{align*} There is no possibility of both events happening.
- \begin{align*}P(A \ or \ B) = P(A) + P(B).\end{align*} To find the probability of either event happening, sum the individual probabilities.
Example A
There are 7 marbles in a bag: 4 green, 2 blue and 1 red. Peter reaches into the bag and blindly picks a single marble. The following letters refer to these events:
- \begin{align*}A\end{align*} – the marble is red
- \begin{align*}B\end{align*} – the marble is blue
- \begin{align*}C\end{align*} – the marble is green
Find the following:
a. \begin{align*}P(A)\end{align*}
b. \begin{align*}P(B)\end{align*}
c. \begin{align*}P(C)\end{align*}
d. \begin{align*}P(B \ or \ A)\end{align*}
e. \begin{align*}P(C \ or \ A)\end{align*}
f. \begin{align*}P(C \ or \ B \ or \ A) \end{align*}
g. \begin{align*}P(A \ and \ C)\end{align*}
Solution
Look at the 3 events \begin{align*}A, B\end{align*} and \begin{align*}C\end{align*}. They must be mutually exclusive: if we pick a single marble from the bag then it must be either red or blue or green. There is no possibility of it being both blue and green!
a. There are 7 marbles and only one is red, so \begin{align*}P(A) = \frac{1}{7}\end{align*}.
b. There are 7 marbles and 2 are blue, so \begin{align*}P(B) = \frac{2}{7}.\end{align*}
c. There are 7 marbles and 4 are green, so \begin{align*}P(C) = \frac{4}{7}\end{align*}.
d. The events are mutually exclusive, therefore \begin{align*}P(B \ or \ A) = P(B) + P(A) = \frac{2}{7} + \frac{1}{7} = \frac{3}{7}\end{align*}.
e. The events are mutually exclusive, therefore \begin{align*} P(C \ or \ A) = P(C) + P(A) = \frac{4}{7} + \frac{1}{7} = \frac{5}{7}\end{align*}.
f. The events are mutually exclusive, therefore \begin{align*}P(C \ or \ B \ or \ A) = P(C) + P(B) + P(A) = \frac{4}{7} + \frac{2}{7} + \frac{1}{7} = \frac{7}{7} =1.\end{align*}
g. The events are mutually exclusive, so \begin{align*}P(A \ and \ C) = 0\end{align*}.
The last two results make sense: \begin{align*}P(C \ or \ B \ or \ A)\end{align*} means the probability of the marble being green blue or red. It must be one of these. And \begin{align*}P(A \ and \ C)\end{align*} is the probability of the marble being both red and green. There are no such marbles in the bag!
Earlier we learned about permutations and combinations. We often have to use these calculations when determining the probability of mutually exclusive events.
Example B
There are 7 marbles in a bag: 4 green, 2 blue and 1 red. Peter reaches into the bag and blindly picks out 4 marbles. Find the probability that he removes at least 3 green marbles.
Solution
There are 2 distinct ways this could occur:
a) Peter picks 3 green marbles and 1 other.
b) Peter picks 4 green marbles.
These events are mutually exclusive – he cannot remove (three green and one other) and (four green) at once. If we find \begin{align*}P(A)\end{align*} and \begin{align*}P(B)\end{align*} we know that the total probability \begin{align*}P(A \ or \ B) = P(A) + P(B)\end{align*}.
We are choosing 4 marbles from a bag containing 7 marbles. The total number of marble combinations there are is \begin{align*}_7C_4 = \frac{7!}{4! 3!} =35\end{align*} possible combinations.
a) The number of combinations that contain 3 green marbles +1 other:
First, the 3 greens. We are choosing 3 green marbles out of a total of 4: \begin{align*}_4C_3 = \frac{4!}{3! 1!} =4\end{align*}
Now to choose 1 other – we’re picking 1 marble out of the 3 non-green ones: \begin{align*}_3C_1 = \frac{3!}{2! 1!} =3\end{align*}
The total number of combinations of 3 greens and 1 other = \begin{align*}_4C_3 \times _3C_1 = 4 \times 3 = 12\end{align*}
So \begin{align*} P(A) = \frac{12}{35}\end{align*}.
b) The number of combinations that contain 4 green balls:
We are choosing 4 from 4 possible green balls: \begin{align*}_4C_4 = 1\end{align*} possible combination
So \begin{align*}P(B) =\frac{1}{35}\end{align*}
\begin{align*}P(A \ or \ B) = P(A) + P(B)\end{align*}, so the probability of getting at least 3 green balls is \begin{align*}\frac{12}{35} + \frac{1}{35} =\frac{13}{35}\end{align*} or slightly better than 1 in 3.
Find the Probability of Overlapping Events
Sometimes we wish to look at overlapping events. In essence, this means two events that are NOT mutually exclusive. For instance, if you pick a card at random from a standard deck, what is the probability that you’ll get a card which is either a seven or a diamond? Let’s use this as our next example:
Example C
If you pick a card at random from a standard 52-card deck, what is the probability that you get a card which is either a seven or a diamond?
Solution
One thing we can say for certain is that these two events are not mutually exclusive (it is possible to get a card which both a seven and a diamond). First of all, let’s look at the information we have:
- There are 52 cards – the chances of picking any particular one is \begin{align*}\frac{1}{52}\end{align*}.
- There are 4 sevens (diamond, heart, club, spade) – the chances of picking a seven is \begin{align*}\frac{4}{52} =\frac{1}{13} \end{align*}.
- There are 13 diamonds (ace through king) – the chances of picking a diamond is \begin{align*}\frac{13}{52} =\frac{1}{4} \end{align*}.
- The chances of picking the seven of diamonds is \begin{align*}\frac{1}{52}\end{align*}.
So there are 4 sevens, 13 diamonds and 1 card that is both (the seven of diamonds). That means we can’t just add up the number of sevens and the number of diamonds, because if we did that, we’d be counting the seven of diamonds twice. Instead, we have to add up the number of sevens and the number of diamonds, and then subtract the number of cards that fit in both categories. That means there are \begin{align*}(4 + 13 - 1) = 16\end{align*} cards that are seven or diamond. The probability of getting a seven or a diamond is therefore \begin{align*}\frac{16}{52}=\frac{4}{13}\end{align*}.
We can check this by listing all the cards in the deck and highlighting those that are sevens or diamonds:
You can see that there are 16 cards that fit, not the 17 we’d get if we just added up the sevens and the diamonds.
Look again at the numbers: the number of cards that are seven or diamond is (number of sevens) plus (number of diamonds) minus (number of seven and diamond). In probability terms we can write this as:
\begin{align*}P(\text{seven or diamonds}) = P(\text{seven}) + P(\text{diamonds}) - P(\text{seven and diamonds})\end{align*}
This leads to a general formula:
\begin{align*}\text{For overlapping events}: P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B)\end{align*}
Watch this video for help with the Examples above.
CK-12 Foundation: Mutually Exclusive Events
Vocabulary
- If our two events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are mutually exclusive:
\begin{align*}P(A \ and \ B) = 0.\end{align*} There is no possibility of both events happening.
\begin{align*}P(A \ or \ B) = P(A) + P(B).\end{align*} To find the probability of either event happening, sum the individual probabilities.
- If our two events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are overlapping:
\begin{align*}P(A \ and \ B) \ne 0\end{align*}
\begin{align*}P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B)\end{align*}
Guided Practice
A cooler contains 6 cans of Sprite, 9 cans of Coke, 4 cans of Dr Pepper and 7 cans of Pepsi. If a can is selected at random, calculate the probability that it is either Pepsi or Coke.
Solution:
Selecting a Pepsi or selecting a Coke are mutually exclusive events. This means we can use the formula \begin{align*}P(A\text{ or }B)=P(A)+P(B).\end{align*}
Finding the probability of each event separately:
\begin{align*}P(\text{Pepsi})=\frac{\text{Cans of Pepsi}}{\text{Total cans of soda}}=\frac{7}{26}\approx 0.269\end{align*}
\begin{align*}P(\text{Coke})=\frac{\text{Cans of Coke}}{\text{Total cans of soda}}=\frac{9}{26}\approx 0.346\end{align*}
This means that the probability of selecting a Pepsi or a Coke is:
\begin{align*}P(\text{Pepsi or Coke})=P(\text{Pepsi})+P(\text{Coke})=0.269+0.346=0.615.\end{align*}
There is a 61.5% probability of selecting a Pepsi or a Coke.
Practice
For 1-6, determine whether the following pairs of events are mutually exclusive or overlapping:
- The next car you see being red; the next car you see being a Ford.
- A train being on time; the train being full.
- Flipping a coin and getting heads; flipping a coin and getting tails.
- Selecting 3 cards and getting an ace; selecting 3 cards and getting a king.
- Selecting 3 cards and getting 2 aces; selecting 3 cards and getting 2 kings.
- A person’s age is an even number; a person’s age is a prime number.
For 7-10, a card is selected at random from a standard 52 card deck. Calculate the probability that:
- The card is either a red card or an even number (2, 4, 6, 8 or 10).
- The card is both a red card and an even number.
- The card is red or even but not both.
- The card is black or red but not an ace.