What are complementary events? In probability theory, the complement of an event *A* is the event not *A*; this complementary event is often denoted *A’* or *A ^{c}*. Let’s illustrate with a few examples.

If our event *A* is “it rains today,” then the complement, *A’*, is the event “it doesn’t rain today.” If you’re drawing a card from a standard 52-card deck, and the event *A* is “you draw a diamond,” then the complement *A’* is “you don’t draw a diamond.”

What is common to these examples is that the event *A* and its complement *A’* are **mutually exclusive** and **exhaustive**. They are mutually exclusive because the two events cannot occur at the same time, and they are exhaustive because the sum of their probabilities must add to 100%.

### Example 1: Complementary events with a standard 6-sided die

Let’s say we are rolling a standard 6-sided die, and our event *A* is “rolls a 5 or 6.” Our complement, *A’*, would then be “rolls a 1, 2, 3, or 4.” We can illustrate this as follows:

Let’s look at the probability of *A* and *A’*.

*P(A) = 2/6
P(A’) = 4/6*

Note that the two events are mutually exclusive (you can’t simultaneously roll a 2 and a 5, for instance) and exhaustive, since the sum of the probabilities above is 1. This is formalized by the **Complement Rule**.

### The Complement Rule

The Complement Rule says that for an event *A* and its complement *A’*, the probability of *A* is equal to one minus the probability of *A’*:

*P(A’) = 1 – P(A)*

This will apply to all events and their complements. Let’s practice, this time with a slightly more advanced example.

### Example 2: Complementary events with multiple coin flips

You flip a coin four times in a row. What is the probability you get at least one heads?

To tackle this problem “head on,” we would need to find out all the possible desired outcomes. Here are a few:

HTTT

HHTT

HTHT

TTHT….

The list goes on. While it’s possible to solve the problem this way, it is tedious and inefficient to do so. We can solve much more quickly by thinking in terms of complements.

If our event *A* is “you get at least heads in four flips”, then the complement *A’* is “you don’t get any heads in four flips,” which is another way of saying “you get all tails.”

Now all we need to do is find the probability of our complement, *A’*, and then subtract this from one.

*P(four tails in a row) = ½*½*½*½ = 1/16*

Therefore, the probability of at least one heads is:

*P(at least one heads) = 1 – P(four tails in a row) = 1 – 1/16 = 15/16 *

There you have it; by harnessing the Complement Rule, a seemingly difficult problem can be solved much more simply.

Do you need more practice with complementary events? Check our statistics blog and videos here!

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