Probability: Complement of an Event

The Complement of an event is the case where the event does not occur. Usually we add a bar over the symbol for an event to denote its complement, so the complement of $A$ is $\bar{A}$.

The probability of the complement of an event is $$P(\bar{A}) = 1 - P(A).$$ Let's say we're curious about whether it will snow tomorrow. We would say that the event of interest $A$ is "It snows tomorrow." Therefore, asking about $P(A)$ means that we want to know "What is the probability that it snows?" If we ask about $P(\bar{A})$, then we want to know "What is the probability of it not snowing tomorrow?" If the weatherman says that there is a $36\%$ chance of snow, then by the logic of complements, there is a $64\%$ chance of not snow - which includes the possibility that the sun will shine, that it will rain or hail, and the possibility of the inevitable heat death of the universe finally occurring. Many other unlikely, but possible states of the weather tomorrow, too.

Seems pretty simple, right? But why does it work?

The probability that something within the realm of possibility happens is always $1$. This is because something will always happen - when you roll a die, one of the sides will always come up (You're not going to let your game be ruined by the fact that the die fell into a crack or off of the table, so you'll pick it up and re-roll it).

Which intuitively makes sense - it is certain (probability = $1$) that some event within the realm of possibility happens, possibly including nothing at all.

Next we use the logical "or" rules in a nifty way. For simplicity, let's assume that all possibilities are disjoint - they cannot occur together, or $P(A \cap B) = 0$. It turns out the property holds even without this claim, but it's easier to see it this way.

As we said before, the probability of absolutely any of the possible events happening is $1$. "Absolutely any event happening" is equivalent to $A$ happens or $B$ happens or $C$ happens... and so on till you have all of the possible events in a giant "or" chain. This means that the probability of all possible events has to add up to $1$.

Let's say there are 4 events, $A$ through $D$

We know $1 = P(A \cup B \cup C \cup D) = P(A) + P(B) + P(C) + P(D)$

We want to know $P(\bar{A})$, which is equivalent to $P(B \cup C \cup D)$ (all the events that are not $A$)

Simple subtraction shows that $1 - P(A) = P(B) + P(C) + P(D) = P(B \cup C \cup D) = P(\bar{A})$

Therefore, $1 - P(A) = P(\bar{A})$

Cool, right?

Complements

No, these aren't compliments, but rather complements. Just in case you're disappointed, you should know that we think you're awesome!

In probability and statistics, the complement of an event is the probability that the event doesn't occur.

In other words, the complement of event A is the probability that event A does not occur.

Here's how we abbreviate the complement in statistics language:

You might be wondering, "How the HECK am I supposed to find the probability of something not happening?"

Luckily, it's a pretty straightforward formula:

Let's start with a simple example.

Let's try another relatively simple example.

To make the game more exciting, I now just have to guess the type of card--either suit or rank (number, ace ($1$), jack, queen, or king).

Right! Since there are $4$ suits in a deck, then the probability of picking a random suit and pulling a card in that suit is $\frac{1}{4}$!

Great! Since there are $13$ ranks in each suit, $4$ suits total, if I guess a random rank, then there is a $\frac{4}{52}$ chance I pull that card! $\frac{4}{52} = \frac{1}{13}$

(Video) Probability of the Complement of an Event

by HCCMathHelp

HCCMathHelp has a great video showing how to find the probability of a complement of an event.

$\large{\text{Summary}}$

The complement of an event is the probability of an event NOT happening. If the probability of an event is $P(A)$, then the complement of $A$ is $\bar{A}$ and therefore the probability of the complement happening is $P(\bar{A})$.

To calculate $\bar{A}$, you only need to use this simple equation: $$P(\bar{A})= 1- P(A)$$

This is because the probability of all of your events must add up to $1$, or $100$%.

The first example in the video uses a question based on airline flights. "An airline reports that $94$% of its flights arrive on time, what is the probability that a flight will be late?"

For this problem we are going to call the chance that the airline is on time $A$. Here, $A$ is $94$% or $A=0.94$. $\bar{A}$ is probability that the plane does not arrive on time, so we can use our equation to find $\bar{A}$.

$$\begin{align}P(\bar{A})&= 1- P(A)\\ P(\bar{A})&= 1 - 0.94\\ P(\bar{A})&= 0.06 \end{align}$$

The chance that the airline is not on time is $6$%.

The second example used shows a spinner with $5$ sections labeled $A$, $B$, $C$, $D$, and $E$. The probability of $A$ is $P(A)=0.23$. He wants to find the probability of the complement of $A$, or $\bar{A}$. Once again you just have to use the equation.

$$\begin{align} P(\bar{A}) &= 1 - P(A)\\ P(\bar{A}) &= 1 - 0.23\\ P(\bar{A}) &= 0.77\\ \end{align}$$

The probability of the spinner landing on not $A$ is $0.77$.

The third example is as follows: "In a bag containing $29$ marbles, $5$ of the marbles are red, and $2$ of the marbles are green. What is the probability of randomly selecting a marble that is not red nor green?"

First we need to find the probability of getting a red or green marble. $P(R)$ will be the probability of getting a red marble, and $P(G)$ will be the probability of getting a green marble.

$$\begin{align} P(R) &= \frac{5}{29}\\ P(G) &= \frac{2}{29}\\ \end{align}$$

Now we're going to add the probability of both events together to get $P(R\cup G)$

$$\begin{align} P(R \cup G) &= \frac{5}{29} + \frac{2}{29}\\ P(R \cup G)&=\frac{7}{29} \end{align}$$

The probability of getting the complement of getting red or green, or the chance that you pull a marble that isn't red or green, is $P(\bar{R \cup G})$. To calculate this, we can use $P(\bar{R \cup G}) = 1 - P(R \cup G)$

$$\begin{align} P(\bar{R \cup G}) &= 1 - P(R \cup G)\\ P(\bar{R \cup G}) &= 1 - \frac{7}{29}\\ P(\bar{R \cup G}) &= \frac{22}{29} \end{align}$$

The probability of not pulling a red or green marble is $\frac{22}{29}$.