# Probability: Complement

Complement of an Event: All outcomes that are **NOT** the event.

When the event is Heads, the complement is Tails | |

When the event is {Monday, Wednesday} the complement is {Tuesday, Thursday, Friday, Saturday, Sunday} | |

When the event is {Hearts} the complement is {Spades, Clubs, Diamonds, Jokers} |

So the Complement of an event is all the **other** outcomes (**not** the ones we want).

And together the Event and its Complement make all possible outcomes.

## Probability

Probability of an event happening = \frac{Number of ways it can happen}{Total number of outcomes}

### Example: the chances of rolling a "4" with a die

**Number of ways it can happen: 1** (there is only 1 face with a "4" on it)

**Total number of outcomes: 6** (there are 6 faces altogether)

So the probability = \frac{1}{6}

The probability of an event is shown using "P":

**P(A)** means "Probability of Event A"

The complement is shown by a little mark after the letter such as A' (or sometimes A^{c} or A):

**P(A')** means "Probability of the complement of Event A"

The two probabilities always add to 1

P(A) + P(A') = 1

### Example: Rolling a "5" or "6"

**Event A** is {5, 6}

Number of ways it can happen: 2

Total number of outcomes: 6

P(A) = \frac{2}{6} = \frac{1}{3}

The **Complement of Event A** is {1, 2, 3, 4}

Number of ways it can happen: 4

Total number of outcomes: 6

P(A') = \frac{4}{6} = \frac{2}{3}

Let us add them:

P(A) + P(A') = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1

Yep, that makes 1

It makes sense, right? **Event A** plus all outcomes that are **not Event A** make up all possible outcomes.

## Why is the Complement Useful?

It is sometimes easier to work out the complement first.

### Example. Throw two dice. What is the probability the two scores are **different**?

Different scores are like getting a **2 and 3**, or a **6 and 1**. It is a long list:

A = { (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,3), (2,4), (1,5), (1,6),

(3,1), (3,2), ... etc ! }

But the complement (which is when the two scores are the same) is only **6 outcomes**:

A' = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }

And its probability is:

P(A') = \frac{6}{36} = \frac{1}{6}

Knowing that P(A) and P(A') together make 1, we can calculate:

P(A) | = 1 − P(A') |

= 1 − \frac{1}{6} | |

= \frac{5}{6} |

So in this case (and many others) it is easier to work out **P(A')** first, then calculate **P(A) = 1 − P(A')**