# 13.7 Probability  (Page 3/18)

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A card is drawn from a standard deck. Find the probability of drawing a red card or an ace.

$\text{\hspace{0.17em}}\frac{7}{13}\text{\hspace{0.17em}}$

## Computing the probability of mutually exclusive events

Suppose the spinner in [link] is spun again, but this time we are interested in the probability of spinning an orange or a $\text{\hspace{0.17em}}d.\text{\hspace{0.17em}}$ There are no sectors that are both orange and contain a $\text{\hspace{0.17em}}d,\text{\hspace{0.17em}}$ so these two events have no outcomes in common. Events are said to be mutually exclusive events    when they have no outcomes in common. Because there is no overlap, there is nothing to subtract, so the general formula is

$\text{\hspace{0.17em}}P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)\text{\hspace{0.17em}}$

Notice that with mutually exclusive events, the intersection of $\text{\hspace{0.17em}}E\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}F\text{\hspace{0.17em}}$ is the empty set. The probability of spinning an orange is $\text{\hspace{0.17em}}\frac{3}{6}=\frac{1}{2}\text{\hspace{0.17em}}$ and the probability of spinning a $d$ is $\text{\hspace{0.17em}}\frac{1}{6}.\text{\hspace{0.17em}}$ We can find the probability of spinning an orange or a $d$ simply by adding the two probabilities.

The probability of spinning an orange or a $d$ is $\text{\hspace{0.17em}}\frac{2}{3}.$

## Probability of the union of mutually exclusive events

The probability of the union of two mutually exclusive events $\text{\hspace{0.17em}}E\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}F\text{\hspace{0.17em}}$ is given by

$\text{\hspace{0.17em}}P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)\text{\hspace{0.17em}}$

Given a set of events, compute the probability of the union of mutually exclusive events.

1. Determine the total number of outcomes for the first event.
2. Find the probability of the first event.
3. Determine the total number of outcomes for the second event.
4. Find the probability of the second event.

## Computing the probability of the union of mutually exclusive events

A card is drawn from a standard deck. Find the probability of drawing a heart or a spade.

The events “drawing a heart” and “drawing a spade” are mutually exclusive because they cannot occur at the same time. The probability of drawing a heart is $\text{\hspace{0.17em}}\frac{1}{4},\text{\hspace{0.17em}}$ and the probability of drawing a spade is also $\text{\hspace{0.17em}}\frac{1}{4},\text{\hspace{0.17em}}$ so the probability of drawing a heart or a spade is

$\text{\hspace{0.17em}}\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\text{\hspace{0.17em}}$

A card is drawn from a standard deck. Find the probability of drawing an ace or a king.

$\text{\hspace{0.17em}}\frac{2}{13}\text{\hspace{0.17em}}$

## Using the complement rule to compute probabilities

We have discussed how to calculate the probability that an event will happen. Sometimes, we are interested in finding the probability that an event will not happen. The complement of an event $\text{\hspace{0.17em}}E,\text{\hspace{0.17em}}$ denoted $\text{\hspace{0.17em}}{E}^{\prime },\text{\hspace{0.17em}}$ is the set of outcomes in the sample space that are not in $\text{\hspace{0.17em}}E.\text{\hspace{0.17em}}$ For example, suppose we are interested in the probability that a horse will lose a race. If event $\text{\hspace{0.17em}}W\text{\hspace{0.17em}}$ is the horse winning the race, then the complement of event $\text{\hspace{0.17em}}W\text{\hspace{0.17em}}$ is the horse losing the race.

To find the probability that the horse loses the race, we need to use the fact that the sum of all probabilities in a probability model must be 1.

$\text{\hspace{0.17em}}P\left({E}^{\prime }\right)=1-P\left(E\right)\text{\hspace{0.17em}}$

The probability of the horse winning added to the probability of the horse losing must be equal to 1. Therefore, if the probability of the horse winning the race is $\text{\hspace{0.17em}}\frac{1}{9},\text{\hspace{0.17em}}$ the probability of the horse losing the race is simply

$\text{\hspace{0.17em}}1-\frac{1}{9}=\frac{8}{9}\text{\hspace{0.17em}}$

## The complement rule

The probability that the complement of an event    will occur is given by

$\text{\hspace{0.17em}}P\left({E}^{\prime }\right)=1-P\left(E\right)\text{\hspace{0.17em}}$

## Using the complement rule to calculate probabilities

Two six-sided number cubes are rolled.

1. Find the probability that the sum of the numbers rolled is less than or equal to 3.
2. Find the probability that the sum of the numbers rolled is greater than 3.

The first step is to identify the sample space, which consists of all the possible outcomes. There are two number cubes, and each number cube has six possible outcomes. Using the Multiplication Principle, we find that there are $6×6,\text{\hspace{0.17em}}$ or total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube.

 $\text{1-1}$ $\text{1-2}$ $\text{1-3}$ $\text{1-4}$ $\text{1-5}$ $\text{1-6}$ $\text{2-1}$ $\text{2-2}$ $\text{2-3}$ $\text{}$ $\text{2-4}$ $\text{2-5}$ $\text{2-6}$ $\text{3-1}$ $\text{3-2}$ $\text{3-3}$ $\text{3-4}$ $\text{3-5}$ $\text{3-6}$ $\text{4-1}$ $\text{4-2}$ $\text{4-3}$ $\text{4-4}$ $\text{4-5}$ $\text{4-6}$ $\text{5-1}$ $\text{5-2}$ $\text{5-3}$ $\text{5-4}$ $\text{5-5}$ $\text{5-6}$ $\text{6-1}$ $\text{6-2}$ $\text{6-3}$ $\text{6-4}$ $\text{6-5}$ $\text{6-6}$
1. We need to count the number of ways to roll a sum of 3 or less. These would include the following outcomes: 1-1, 1-2, and 2-1. So there are only three ways to roll a sum of 3 or less. The probability is
$\text{\hspace{0.17em}}\frac{3}{36}=\frac{1}{12}\text{\hspace{0.17em}}$
2. Rather than listing all the possibilities, we can use the Complement Rule. Because we have already found the probability of the complement of this event, we can simply subtract that probability from 1 to find the probability that the sum of the numbers rolled is greater than 3.

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