### Learning Outcomes

- Calculate the probability of a complementary event
- Calculate the probability of two independent events both occurring together
- Calculate the probability of two mutually exclusive events
- Calculate the probability of two events that are not mutually exclusive

### Recall operations on fractions

Adding and subtracting fractions with common denominators

[latex]\dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm b}{c}[/latex]

In the two equations below, note that this relationship is described in both directions.

That is, it is also true that

[latex]\dfrac{a\pm b}{c}=\dfrac{a}{c}\pm \dfrac{b}{c}[/latex]

The second equation furthermore includes the fact that

[latex]\dfrac{a}{a}=1[/latex]

## Complementary Events

Now let us examine the probability that an event does **not** happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a six: the answer is *P*(six) =1/6. Now consider the probability that we do *not* roll a six: there are 5 outcomes that are not a six, so the answer is *P*(not a six) = [latex]\frac{5}{6}[/latex]. Notice that

[latex]P(\text{six})+P(\text{not a six})=\frac{1}{6}+\frac{5}{6}=\frac{6}{6}=1[/latex]

This is not a coincidence. Consider a generic situation with *n* possible outcomes and an event *E* that corresponds to *m* of these outcomes. Then the remaining *n* – *m* outcomes correspond to *E* not happening, thus

[latex]P(\text{not}E)=\frac{n-m}{n}=\frac{n}{n}-\frac{m}{n}=1-\frac{m}{n}=1-P(E)[/latex]

### Complement of an Event

The **complement** of an event is the event “*E* doesn’t happen”

- The notation [latex]\bar{E}[/latex] is used for the complement of event
*E*. - We can compute the probability of the complement using [latex]P\left({\bar{E}}\right)=1-P(E)[/latex]
- Notice also that [latex]P(E)=1-P\left({\bar{E}}\right)[/latex]

### example

If you pull a random card from a deck of playing cards, what is the probability it is not a heart?

This situation is explained in the following video.

### Try It

## Probability of two independent events

### example

Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin and a 6 on the die.

The prior example contained two *independent* events. Getting a certain outcome from rolling a die had no influence on the outcome from flipping the coin.

### Independent Events

Events A and B are **independent events** if the probability of Event B occurring is the same whether or not Event A occurs.

### example

Are these events independent?

- A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.
- The two events (1) “It will rain tomorrow in Houston” and (2) “It will rain tomorrow in Galveston” (a city near Houston).
- You draw a card from a deck, then draw a second card without replacing the first.

When two events are independent, the probability of both occurring is the product of the probabilities of the individual events.

### recall multiplying fractions

To multiply fractions, place the product of the numerators over the product of the denominators.

[latex]\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{ac}{bd}[/latex]

*P*(*A* and *B*) for independent events

If events *A* and *B* are independent, then the probability of both *A* and *B *occurring is

[latex]P\left(A\text{ and }B\right)=P\left(A\right)\cdot{P}\left(B\right)[/latex]

where *P*(*A* and *B*) is the probability of events *A* and *B* both occurring, *P*(*A*) is the probability of event *A* occurring, and *P*(*B*) is the probability of event *B* occurring

If you look back at the coin and die example from earlier, you can see how the number of outcomes of the first event multiplied by the number of outcomes in the second event multiplied to equal the total number of possible outcomes in the combined event.

### Recall fraction reduction

To write a fraction in reduced terms, first take the prime factorization of the numerator and denominator, then cancel out factors that are common in the numerator and the denominator.

Ex. [latex]\dfrac{12}{18}=\dfrac{\cancel{2}\cdot 2\cdot \cancel{3}}{\cancel{2}\cdot 3\cdot \cancel{3}}=\dfrac{2}{3}[/latex]

### example

In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you randomly reach in and pull out a pair of socks and a tee shirt, what is the probability both are white?

Examples of joint probabilities are discussed in this video.

### Try It

The previous examples looked at the probability of *both* events occurring. Now we will look at the probability of *either* event occurring.

### example

Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin *or* a 6 on the die.

*P*(*A* or *B*)

The probability of either *A* or *B *occurring (or both) is

[latex]P(A\text{ or }B)=P(A)+P(B)–P(A\text{ and }B)[/latex]

### example

Suppose we draw one card from a standard deck. What is the probability that we get a Queen or a King?

See more about this example and the previous one in the following video.

In the last example, the events were **mutually exclusive**, so *P*(*A* or *B*) = *P*(*A*) + *P*(*B*).

### Try It

### example

Suppose we draw one card from a standard deck. What is the probability that we get a red card or a King?

### Try It

In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you reach in and randomly grab a pair of socks and a tee shirt, what the probability at least one is white?

### Example

The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person:

- Has a red car
*and*got a speeding ticket - Has a red car
*or*got a speeding ticket.

Speeding ticket | No speeding ticket | Total | |

Red car | 15 | 135 | 150 |

Not red car | 45 | 470 | 515 |

Total | 60 | 605 | 665 |

This table example is detailed in the following explanatory video.