# Probability for Multiple Events

### Learning Outcomes

- Find the probability of a union of two events.
- Find the probability of two events that share no common outcomes.
- Find the probability that an event will not happen.
- Find the number of events in a sample space that that includes many choices.

**union of two events**[latex]E\text{ and }F,\text{written }E\cup F[/latex], is the event that occurs if either or both events occur.

[latex]P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)-P\left(E\cap F\right)[/latex]

Suppose the spinner below is spun. We want to find the probability of spinning orange or spinning a [latex]b[/latex]. There are a total of 6 sections, and 3 of them are orange. So the probability of spinning orange is [latex]\frac{3}{6}=\frac{1}{2}[/latex]. There are a total of 6 sections, and 2 of them have a [latex]b[/latex]. So the probability of spinning a [latex]b[/latex] is [latex]\frac{2}{6}=\frac{1}{3}[/latex]. If we added these two probabilities, we would be counting the sector that is both orange and a [latex]b[/latex] twice. To find the probability of spinning an orange or a [latex]b[/latex], we need to subtract the probability that the sector is both orange and has a [latex]b[/latex].[latex]\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}=\dfrac{2}{3}[/latex]

The probability of spinning orange or a [latex]b[/latex] is [latex]\dfrac{2}{3}[/latex].### A General Note: Probability of the Union of Two Events

The probability of the union of two events [latex]E[/latex] and [latex]F[/latex] (written [latex]E\cup F[/latex] ) equals the sum of the probability of [latex]E[/latex] and the probability of [latex]F[/latex] minus the probability of [latex]E[/latex] and [latex]F[/latex] occurring together [latex]\text{(}[/latex] which is called the**intersection**of [latex]E[/latex] and [latex]F[/latex] and is written as [latex]E\cap F[/latex] ). [latex-display]P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)-P\left(E\cap F\right)[/latex-display]

### tip for success

The*union symbol*given above [latex]\cup[/latex] is the same symbol you used in the past to express the union of two intervals. Here we use it to represent the union of two events. Mathematically, it represents the word

*or.*

The union of *E *and *F* includes all the elements that could be present in *E or in F, one or the other.*

*intersection symbol*given above [latex]\cap[/latex] may be new notation for you. It is used to express the intersection of events. Mathematically, it represents the word

*and*.

The intersection of *E* and *F* includes all the elements present in *both E and F, and not in just one or the other.*

### Example: Computing the Probability of the Union of Two Events

A card is drawn from a standard deck. Find the probability of drawing a heart or a 7.Answer: A standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability of drawing a heart is [latex]\frac{1}{4}[/latex]. There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing a 7 is [latex]\frac{1}{13}[/latex]. The only card in the deck that is both a heart and a 7 is the 7 of hearts, so the probability of drawing both a heart and a 7 is [latex]\frac{1}{52}[/latex]. Substitute [latex]P\left(H\right)=\dfrac{1}{4}, P\left(7\right)=\dfrac{1}{13}, \text{and} P\left(H\cap 7\right)=\dfrac{1}{52}[/latex] into the formula.

[latex]\begin{align}P\left(E\cup F\right)&=P\left(E\right)+P\left(F\right)-P\left(E\cap F\right) \\ &=\frac{1}{4}+\frac{1}{13}-\frac{1}{52} \\ &=\frac{4}{13} \end{align}[/latex]

The probability of drawing a heart or a 7 is [latex]\dfrac{4}{13}[/latex].### Try It

A card is drawn from a standard deck. Find the probability of drawing a red card or an ace.Answer: [latex-display]\dfrac{7}{13}[/latex-display]

## Computing the Probability of Mutually Exclusive Events

Suppose the spinner from earlier is spun again, but this time we are interested in the probability of spinning an orange or a [latex]d[/latex]. There are no sectors that are both orange and contain a [latex]d[/latex], 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[latex]P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)[/latex]

Notice that with mutually exclusive events, the intersection of [latex]E[/latex] and [latex]F[/latex] is the empty set. The probability of spinning an orange is [latex]\frac{3}{6}=\frac{1}{2}[/latex] and the probability of spinning a [latex]d[/latex] is [latex]\frac{1}{6}[/latex]. We can find the probability of spinning an orange or a [latex]d[/latex] simply by adding the two probabilities.[latex]\begin{align}P\left(E\cup F\right)&=P\left(E\right)+P\left(F\right) \\ &=\frac{1}{2}+\frac{1}{6} \\ &=\frac{2}{3} \end{align}[/latex]

The probability of spinning an orange or a [latex]d[/latex] is [latex]\dfrac{2}{3}[/latex].### A General Note: Probability of the Union of Mutually Exclusive Events

The probability of the union of two*mutually exclusive*events [latex]E[/latex] and [latex]F[/latex] is given by

[latex]P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)[/latex]

### How To: Given a set of events, compute the probability of the union of mutually exclusive events.

- Determine the total number of outcomes for the first event.
- Find the probability of the first event.
- Determine the total number of outcomes for the second event.
- Find the probability of the second event.
- Add the probabilities.

### Example: 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.Answer: 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 [latex]\frac{1}{4}[/latex], and the probability of drawing a spade is also [latex]\frac{1}{4}[/latex], so the probability of drawing a heart or a spade is

[latex]\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}[/latex]

### Try It

A card is drawn from a standard deck. Find the probability of drawing an ace or a king.Answer: [latex-display]\dfrac{2}{13}[/latex-display]

## Find the Probability That an Even Will Not Happen

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**[latex]E[/latex], denoted [latex]{E}^{\prime }[/latex], is the set of outcomes in the sample space that are not in [latex]E[/latex]. For example, suppose we are interested in the probability that a horse will lose a race. If event [latex]W[/latex] is the horse winning the race, then the complement of event [latex]W[/latex] 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.

[latex]P\left({E}^{\prime }\right)=1-P\left(E\right)[/latex]

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 [latex]\frac{1}{9}[/latex], the probability of the horse losing the race is simply[latex]1-\dfrac{1}{9}=\dfrac{8}{9}[/latex]

### A General Note: The Complement Rule

The probability that the**complement of an event**will occur is given by [latex-display]P\left({E}^{\prime }\right)=1-P\left(E\right)[/latex-display]

### Example: Using the Complement Rule to Calculate Probabilities

Two six-sided number cubes are rolled.- Find the probability that the sum of the numbers rolled is less than or equal to 3.
- Find the probability that the sum of the numbers rolled is greater than 3.

Answer: 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 [latex]6\times 6[/latex], or [latex]\text{ 36 }[/latex] total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube.

[latex]\text{1 - 1}[/latex] | [latex]\text{1 - 2}[/latex] | [latex]\text{1 - 3}[/latex] | [latex]\text{1 - 4}[/latex] | [latex]\text{1 - 5}[/latex] | [latex]\text{1 - 6}[/latex] |

[latex]\text{2 - 1}[/latex] | [latex]\text{2 - 2}[/latex] | [latex]\text{2 - 3}[/latex] | [latex]\text{}[/latex] [latex]\text{2 - 4}[/latex] | [latex]\text{2 - 5}[/latex] | [latex]\text{2 - 6}[/latex] |

[latex]\text{3 - 1}[/latex] | [latex]\text{3 - 2}[/latex] | [latex]\text{3 - 3}[/latex] | [latex]\text{3 - 4}[/latex] | [latex]\text{3 - 5}[/latex] | [latex]\text{3 - 6}[/latex] |

[latex]\text{4 - 1}[/latex] | [latex]\text{4 - 2}[/latex] | [latex]\text{4 - 3}[/latex] | [latex]\text{4 - 4}[/latex] | [latex]\text{4 - 5}[/latex] | [latex]\text{4 - 6}[/latex] |

[latex]\text{5 - 1}[/latex] | [latex]\text{5 - 2}[/latex] | [latex]\text{5 - 3}[/latex] | [latex]\text{5 - 4}[/latex] | [latex]\text{5 - 5}[/latex] | [latex]\text{5 - 6}[/latex] |

[latex]\text{6 - 1}[/latex] | [latex]\text{6 - 2}[/latex] | [latex]\text{6 - 3}[/latex] | [latex]\text{6 - 4}[/latex] | [latex]\text{6 - 5}[/latex] | [latex]\text{6 - 6}[/latex] |

- 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
[latex]\dfrac{3}{36}=\dfrac{1}{12}[/latex]
- 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.
[latex]\begin{align}P\left({E}^{\prime }\right)&=1-P\left(E\right) \\ &=1-\frac{1}{12} \\ &=\frac{11}{12} \end{align}[/latex]

### Try It

Two number cubes are rolled. Use the Complement Rule to find the probability that the sum is less than 10.Answer: [latex-display]\dfrac{5}{6}[/latex-display]

[embed]## Computing Probability Using Counting Theory

Many interesting probability problems involve counting principles, permutations, and combinations. In these problems, we will use permutations and combinations to find the number of elements in events and sample spaces. These problems can be complicated, but they can be made easier by breaking them down into smaller counting problems. Assume, for example, that a store has 8 cellular phones and that 3 of those are defective. We might want to find the probability that a couple purchasing 2 phones receives 2 phones that are not defective. To solve this problem, we need to calculate all of the ways to select 2 phones that are not defective as well as all of the ways to select 2 phones. There are 5 phones that are not defective, so there are [latex]C\left(5,2\right)[/latex] ways to select 2 phones that are not defective. There are 8 phones, so there are [latex]C\left(8,2\right)[/latex] ways to select 2 phones. The probability of selecting 2 phones that are not defective is:[latex]\begin{align}\frac{\text{ways to select 2 phones that are not defective}}{\text{ways to select 2 phones}}&=\frac{C\left(5,2\right)}{C\left(8,2\right)} \\[1mm] &=\frac{10}{28} \\[1mm] &=\frac{5}{14} \end{align}[/latex]

### tip for success

The following are challenging problems. Work them out on paper as many times as you need to gain insight. You can click to Try Another Version of This Question in the green Try It box as needed. Don't be discouraged if it takes some time!### Example: Computing Probability Using Counting Theory

A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears.- Find the probability that only bears are chosen.
- Find the probability that 2 bears and 3 dogs are chosen.
- Find the probability that at least 2 dogs are chosen.

Answer:

- We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are [latex]C\left(6,5\right)[/latex] ways to choose 5 bears. There are 14 toys, so there are [latex]C\left(14,5\right)[/latex] ways to choose any 5 toys.
[latex]\dfrac{C\left(6\text{,}5\right)}{C\left(14\text{,}5\right)}=\dfrac{6}{2\text{,}002}=\dfrac{3}{1\text{,}001}[/latex]
- We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are [latex]C\left(6,2\right)[/latex] ways to choose 2 bears. There are 5 dogs, so there are [latex]C\left(5,3\right)[/latex] ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are [latex]C\left(6,2\right)\cdot C\left(5,3\right)[/latex] ways to choose 2 bears and 3 dogs. We can use this result to find the probability.
[latex]\dfrac{C\left(6\text{,}2\right)C\left(5\text{,}3\right)}{C\left(14\text{,}5\right)}=\dfrac{15\cdot 10}{2\text{,}002}=\dfrac{75}{1\text{,}001}[/latex]
- It is often easiest to solve "at least" problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or 1 dog could be chosen. When no dogs are chosen, all 5 toys come from the 9 toys that are not dogs. There are [latex]C\left(9,5\right)[/latex] ways to choose toys from the 9 toys that are not dogs. Since there are 14 toys, there are [latex]C\left(14,5\right)[/latex] ways to choose the 5 toys from all of the toys.
[latex]\dfrac{C\left(9\text{,}5\right)}{C\left(14\text{,}5\right)}=\dfrac{63}{1\text{,}001}[/latex]If there is 1 dog chosen, then 4 toys must come from the 9 toys that are not dogs, and 1 must come from the 5 dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are [latex]C\left(5,1\right)\cdot C\left(9,4\right)[/latex] ways to choose 1 dog and 1 other toy.[latex]\dfrac{C\left(5\text{,}1\right)C\left(9\text{,}4\right)}{C\left(14\text{,}5\right)}=\dfrac{5\cdot 126}{2\text{,}002}=\dfrac{315}{1\text{,}001}[/latex]Because these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than 2 dogs are chosen.[latex]\dfrac{63}{1\text{,}001}+\dfrac{315}{1\text{,}001}=\dfrac{378}{1\text{,}001}[/latex]We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen.[latex]1-\dfrac{378}{1\text{,}001}=\dfrac{623}{1\text{,}001}[/latex]

### Try It

A child randomly selects 3 gumballs from a container holding 4 purple gumballs, 8 yellow gumballs, and 2 green gumballs.- Find the probability that all 3 gumballs selected are purple.
- Find the probability that no yellow gumballs are selected.
- Find the probability that at least 1 yellow gumball is selected.

Answer: [latex-display]\begin{align} &\text{a}\text{. }\frac{1}{91} \\[1mm] & \text{b}\text{. }\frac{\text{5}}{\text{91}} \\[1mm] & \text{c}\text{. }\frac{86}{91} \end{align}[/latex-display]

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