# 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** [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.

### A General Note: The Complement Rule

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

### Example 5: 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.

### Solution

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]\frac{3}{36}=\frac{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{array}{l}P\left({E}^{\prime }\right)=1-P\left(E\right)\hfill \\ \text{ }=1-\frac{1}{12}\hfill \\ \text{ }=\frac{11}{12}\hfill \end{array}[/latex]

### Try It 5

Two number cubes are rolled. Use the Complement Rule to find the probability that the sum is less than 10. Solution## Licenses & Attributions

### CC licensed content, Specific attribution

- Precalculus.
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