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Mutually Exclusive Events

Probability of two events that cannot occur at the same time P(A or B) = P(A) + P(B)

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Mutually Exclusive Events

Suppose you select a blueberry muffin for breakfast at your local bakery. Ten minutes later, after you've gobbled down the muffin, you're still hungry so next you select something a little healthier than the muffin, maybe a whole grain bagel to take for the road. Do you think these events are mutually exclusive or mutually inclusive?

Guidance

Two events A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} that cannot occur at the same time are mutually exclusive events. When we add probability calculations of events described by this term, we can apply the word or ()\begin{align*}(\cup)\end{align*}. Mutually exclusive events have no common outcomes. See in the diagram below that P(A and B)=0\begin{align*}P(A \ \text{and} \ B) = 0\end{align*}. Notice that there is no intersection between the possible outcomes of event A\begin{align*}A\end{align*} and the possible outcomes of event B\begin{align*}B\end{align*}. For example, if you were asked to pick a number between 1 and 10, you cannot pick a number that is both even and odd. These events are mutually exclusive.

What if we said that we were choosing a card from a deck of cards? Suppose event A\begin{align*}A\end{align*} is choosing an eight and event B\begin{align*}B\end{align*} is choosing an Ace.

Notice that the sets containing the possible outcomes of the events have no elements in common. Therefore, the events are mutually exclusive.

Example A

What is the probability of randomly picking a number from 1 to 10 that is even or randomly picking a number from 1 to 10 that is odd?

To calculate the probability of picking a number from 1 to 10 that is even or picking a number from 1 to 10 that is odd, you would follow the steps below:

A={2,4,6,8,10}P(A)=510B={1,3,5,7,9}P(B)=510P(A or B)=510+510P(A or B)=1010P(A or B)=1\begin{align*}A = \left \{2, 4, 6, 8, 10 \right \}\\ \\ P(A) = \frac{5}{10}\\ \\ B = \left \{1, 3, 5, 7, 9 \right \}\\ \\ P(B) = \frac{5}{10}\\ \\ P(A \ \text{or} \ B) = \frac{5}{10} + \frac{5}{10}\\ \\ P(A \ \text{or} \ B) = \frac{10}{10}\\ \\ P(A \ \text{or} \ B) = 1\end{align*}

The probability of picking a number from 1 to 10 that is even and picking a number from 1 to 10 that is odd would just be 0, since these are mutually exclusive events. In other words, P(A and B)=0\begin{align*}P(A \ \text{and} \ B) = 0\end{align*}.

Example B

2 fair dice are rolled. What is the probability of getting a sum less than 7 or a sum equal to 10?

P(A)=probability of obtaining a sum less than 7P(A)=1536\begin{align*}P(A) = \text{probability of obtaining a sum less than 7}\\ \\ P(A)=\frac{15}{36}\end{align*}

P(B)=probability of obtaining a sum equal to 10P(B)=336\begin{align*}P(B) = \text{probability of obtaining a sum equal to 10}\\ \\ P(B) = \frac{3}{36}\end{align*}

There are no elements that are common, so the events are mutually exclusive.

P(A or B)P(AB)P(AB)P(AB)P(AB)P(A and B)=P(A)+P(B)=P(A)+P(B)=1536+336=1836=12=0\begin{align*}P(A \ \text{or} \ B) &= P(A) + P(B)\\ P(A \cup B) &= P(A) + P(B)\\ P(A \cup B) &= \frac{15}{36} + \frac{3}{36}\\ P(A \cup B) &= \frac{18}{36}\\ P(A \cup B) &= \frac{1}{2}\\ P(A \ \text{and} \ B) &= 0\end{align*}

Recall that the sum cannot be both less than 7 and equal to 10 at the same time, resulting in
P(A ∩ B)=0

Example C

A card is chosen at random from a standard deck of cards. What is the probability that the card chosen is a diamond or club? Are these events mutually exclusive?

A standard deck of cards contains 52 cards, with 13 hearts, 13 diamonds, 13 spades, and 13 clubs. Since a card cannot be a diamond and a club at the same time, choosing a diamond and choosing a club are mutually exclusive events. Suppose that event A\begin{align*}A\end{align*} is choosing a diamond and event B\begin{align*}B\end{align*} is choosing a club. The probability that the card chosen is a diamond or club can then be calculated as follows:

P(A)=1352P(B)=1352\begin{align*}P(A)=\frac{13}{52}\\ P(B)=\frac{13}{52}\end{align*}

P(A or B)P(AB)P(AB)P(AB)P(AB)P(A and B)=P(A)+P(B)=P(A)+P(B)=1352+1352=2652=12=0\begin{align*}P(A \ \text{or} \ B) &= P(A) + P(B)\\ P(A \cup B) &= P(A) + P(B)\\ P(A \cup B) &= \frac{13}{52} + \frac{13}{52}\\ P(A \cup B) &= \frac{26}{52}\\ P(A \cup B) &= \frac{1}{2}\\ P(A \ \text{and} \ B) &= 0\end{align*}

Points to Consider

• Can mutually exclusive events be independent? Can they be dependent?

Vocabulary

Mutually exclusive events are 2 events that cannot both occur simultaneously. The union of 2 events is represented by \begin{align*}\cup\end{align*}, where the sample space contains 2 events, A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, and each member of the set belongs to A\begin{align*}A\end{align*} or B\begin{align*}B\end{align*}.

Guided Practice

3 coins are tossed simultaneously. What is the probability of getting 1 or 2 heads? Are these events mutually exclusive?

Solution:

When tossing 3 coins simultaneously, there are 23=8\begin{align*}2^3=8\end{align*} possible outcomes. These outcomes are as follows, where H\begin{align*}H\end{align*} represents heads and T\begin{align*}T\end{align*} represents tails:

HHHHTHHHTHTTTHHTHTTTHTTT\begin{align*}HHH\\ HTH\\ HHT\\ HTT\\ THH\\ THT\\ TTH\\ TTT\end{align*}

It's apparent from the list of possible outcomes that there are 3 ways to get 1 head and 3 ways to get 2 heads. Since getting 1 head and getting 2 heads cannot occur at the same time, these events are mutually exclusive. Suppose that event A\begin{align*}A\end{align*} is getting 1 head and event B\begin{align*}B\end{align*} is getting 2 heads. The probability of getting 1 or 2 heads can then be calculated as follows:

P(A)=38P(B)=38\begin{align*}P(A)=\frac{3}{8}\\ P(B)=\frac{3}{8}\end{align*}

P(A or B)P(AB)P(AB)P(AB)P(AB)P(A and B)=P(A)+P(B)=P(A)+P(B)=38+38=68=34=0\begin{align*}P(A \ \text{or} \ B) &= P(A) + P(B)\\ P(A \cup B) &= P(A) + P(B)\\ P(A \cup B) &= \frac{3}{8} + \frac{3}{8}\\ P(A \cup B) &= \frac{6}{8}\\ P(A \cup B) &= \frac{3}{4}\\ P(A \ \text{and} \ B) &= 0\end{align*}

Practice

1. 2 dice are tossed. What is the probability of obtaining a sum equal to 6?
2. 2 dice are tossed. What is the probability of obtaining a sum less than 6?
3. 2 dice are tossed. What is the probability of obtaining a sum greater than 6?
4. 2 dice are tossed. What is the probability of obtaining a sum of at least 6?
5. Thomas bought a bag of jelly beans that contained 10 red jelly beans, 15 blue jelly beans, and 12 green jelly beans. What is the probability of Thomas reaching into the bag and pulling out a blue or green jelly bean?
6. A card is chosen at random from a standard deck of cards. What is the probability that the card chosen is a heart or spade? Are these events mutually exclusive?
7. 3 coins are tossed simultaneously. What is the probability of getting 3 heads or 3 tails? Are these events mutually exclusive?
8. In question 7, what is the probability of getting 3 heads and 3 tails when tossing the 3 coins simultaneously?
9. Are randomly choosing a person who is left-handed and randomly choosing a person who is right-handed mutually exclusive events? Explain your answer.
10. Suppose 2 events are mutually exclusive events. If one of the events is randomly choosing a boy from the freshman class of a high school, what could the other event be? Explain your answer.

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