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12.6: Complement Rule for Probability

Difficulty Level: At Grade Created by: CK-12
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Freddy wants to be a part of the school newspaper staff.  He is hoping the meetings are either on Monday or Wednesday since he has soccer practice on all the other days.  What is the probability that the meetings will not be scheduled on Tuesday, Thursday, Friday, Saturday, or Sunday?

In this concept, you will learn about the Complementary Rule and how to calculate the probability of a complementary event occurring.

Complement Rule for Probability

When one of two disjoint events must occur, the two events are said to be complementary.  Since one or the other event must occur, the sum of the probabilities of the two complementary events adds up to 1, or 100 percent of the outcomes of the events.

For example, the probability of the spinner below landing on either red or blue is 1.  The arrow will land on either red or blue 100 percent of the time.

\begin{align*}P (\text{red or blue}) &= P (\text{red}) + P (\text{blue})\\ &= \frac{1}{2} + \frac{1}{2} = 1\end{align*}

Here are some situations which involve complementary events.

  • Flipping a coin heads or flipping a coin tails.
  • Turning on a light switch on or turning a light switch off.
  • Locking a door or unlocking a door.

Although some complementary events are 50-50 events, such as flipping a coin, not all are.

For example, for the spinner shown:

\begin{align*}P (\text{blue or yellow}) &= P (\text{blue}) + P (\text{yellow})\\ &= \frac{3}{4} + \frac{1}{4}\\ &= 1\end{align*}

The events \begin{align*}B (\text{blue})\end{align*} and \begin{align*}Y (\text{yellow})\end{align*} are complementary because their probabilities add up to 1. But the two complements are not equal in size.

Note that some disjoint events are NOT complementary events. Here, \begin{align*}R (\text{red})\end{align*} and \begin{align*}B (\text{blue})\end{align*} are disjoint events. However, their probabilities do NOT add up to 1 or 100 percent:

\begin{align*}P (\text{red or blue}) &= P (\text{red}) + P (\text{blue})\\ &= \frac{1}{4} + \frac{1}{4}\\ &= \frac{1}{2}\end{align*}

Since the sum of any two complements is 1, if you know the probability of one complement, you can find the probability of the other.

For events \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, suppose the probability of \begin{align*}B\end{align*} is 0.4. That means:

\begin{align*}P (A) + P (B) &= 1\\ P (A) + 0.4 &= 1\end{align*}

Therefore, the probability of \begin{align*}P(A)\end{align*} is 0.6, because:

\begin{align*}P (A) + P (B) &= 1\\ 0.6 + 0.4 &= 1\end{align*}

The Complement Rule states that for any two complements, \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, the value of \begin{align*}P (A) = 1 - P (B)\end{align*}.  In other words, subtract the complement you know from 1 to find the unknown complement.

\begin{align*}A\end{align*} and \begin{align*}B\end{align*} are complements. \begin{align*}P (B) = 0.3\end{align*}. Find \begin{align*}P (A)\end{align*}.

To figure this out, subtract the complement you know, 0.3, from 1 to find \begin{align*}P (B)\end{align*}

\begin{align*}P (B) &= 1 - P (A)\\ &= 1 - 0.3\\ &= 0.7\end{align*}

What is the probability that the arrow will land on red, green, or yellow?

The events are disjoint so the probability of one of them occurring is the sum of their individual probabilities.

\begin{align*}P (\text{red or blue or green}) &= P (\text{red}) + P (\text{blue}) + P (\text{green})\\ &= \frac{1}{4} + \frac{1}{4} + \frac{1}{4}\\ &= \frac{3}{4}\end{align*}

The probability of the complementary event of the spinner landing on yellow is:

\begin{align*}P (\text{yellow}) &= 1 - P (\text{red or blue or green})\\ &=\ 1 - \frac{3}{4}\\ &= \frac{1}{4}\end{align*}

The probability of the Mets winning tonight’s game is 0.6. Predict how likely it is for the Mets to lose tonight’s game.

Winning the game and losing the game are complementary events. 

First, substitute the value for the known probability into the formula for complementary events.

\begin{align*}P (\text{lose}) &= 1 - P (\text{win})\\ &= 1 - 0.6\\\end{align*}

Next, subtract the known probability from 1.

\begin{align*}P (\text{lose}) &= 1 - P (\text{win})\\ &= 1 - 0.6\\ &= 0.4\end{align*}

Then, state the answer as the probability for the complementary event.

The answer is the probability that the Mets will lose tonight's game is 0.4.  In other words, there is a 40% chance that the Mets will lose the game tonight.

Examples

Example 1

Earlier, you were given a problem about Freddy's school newspaper club.

If his soccer practice occurs on Tuesdays, Thursdays, Fridays, Saturdays, and Sundays, what is the probability that the newspaper club meetings will not fall on a soccer practice day?

First, substitute the value for the known probability into the formula for complementary events.  We know that Freddy has soccer practice on 5 out of 7 days:\begin{align*}P (\text{non-soccer practice days}) &= 1 - P (\text{soccer practice days})\\ &= 1 - \frac{5}{7}\\\end{align*}Next, subtract the known probability from 1:\begin{align*}P (\text{non-soccer practice days}) &= 1 - P (\text{soccer practice days})\\ &= 1 - \frac{5}{7}\\ &= \frac{2}{7}\end{align*}Then, state the answer as the probability for the complementary event:

The answer is the probability that the newspaper club will decide to meet on a day when soccer practice is not taking place is  \begin{align*}\frac{2}{7}\end{align*}, or  29%. 

Example 2

Y and Z are complements. If the probability of Y occurring is 14%, what is the probability of Z occurring?

First, substitute the value for the known probability into the formula for complementary events.  In this case, since, the known probability is reported in percentage, we will subtract from 100 instead of 1:\begin{align*}P (\text{Z}) &= 100\ percent - P (\text{Y})\\ &= 100\ percent- 14\ percent\\\end{align*} Next, subtract the known probability from 100 percent:\begin{align*}P (\text{Z}) &= 100\ percent - P (\text{Y})\\ &= 100\ percent - 14\ percent\\ &= 86\ percent\end{align*}Then, state the answer as the probability for the complementary event:

The answer is the probability that Z will occur is 86%.  

Example 3

A and C are complements. If C is .67, find A.

First, substitute the value for the known probability into the formula for complementary events:\begin{align*}P (\text{A}) &= 100\ percent - P (\text{C})\\ &= 100\ percent - 67\ percent\\\end{align*}Next, subtract the known probability from 100 percent:\begin{align*}P (\text{A}) &= 100\ percent - P (\text{C})\\ &= 100\ percent - 67\ percent\\ &= 33\ percent\end{align*}Then, state the answer as the probability for the complementary event:

The answer is the probability that  A will occur is 33%.  

Example 4

If the Yankees have a 45% chance of winning tonight, what is the probability that they won't win?

First, substitute the value for the known probability into the formula for complementary events.  In this case, since, the known probability is reported in percentage, we will subtract from 100 instead of 1:\begin{align*}P (\text{won't win}) &= 100\ percent - P (\text{win})\\ &= 100\ percent- 45\ percent\\\end{align*}Next, subtract the known probability from 100 percent:\begin{align*}P (\text{won't win}) &= 100\ percent - P (\text{win})\\ &= 100\ percent - 45\ percent\\ &= 55\ percent\end{align*}Then, state the answer as the probability for the complementary event:

The answer is the probability that the Yankees won't win is 55%.  

Example 5

D and E are complements. If the probabiity of D is 0.02, what is E?

First, substitute the value for the known probability into the formula for complementary events:\begin{align*}P (\text{E}) &= 1\ - P (\text{D})\\ &= 1- 0.2\\\end{align*}Next, subtract the known probability from 1:\begin{align*}P (\text{E}) &= 1 - P (\text{D})\\ &= 1- 0.2\\ &= 0.8\end{align*}Then, state the answer as the probability for the complementary event:

The answer is the probability that E will occur is 0.8, or 80%.  

Review

Find the complement.

  1. \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are complements. \begin{align*}P (B) = 0.15\end{align*}. Find \begin{align*}P (A)\end{align*}.
  2. \begin{align*}C\end{align*} and \begin{align*}D\end{align*} are complements. \begin{align*}P (C) = 0.8\end{align*}. Find \begin{align*}P (D)\end{align*}.
  3. \begin{align*}G\end{align*} and \begin{align*}H\end{align*} are complements. \begin{align*}P (H) = 49\%\end{align*}. Find \begin{align*}P (G)\end{align*}.
  4. \begin{align*}T\end{align*} and \begin{align*}S\end{align*} are complements. \begin{align*}P (T) = \frac{3}{8}\end{align*}. Find \begin{align*}P (S)\end{align*}.
  5. \begin{align*}L\end{align*} and \begin{align*}K\end{align*} are complements. \begin{align*}P (K) = 0.07\end{align*}. Find \begin{align*}P (L)\end{align*}.
  6. \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are complements. \begin{align*}P (B) = 0.125\end{align*}. Find \begin{align*}P (A)\end{align*}.
  7. \begin{align*}N\end{align*} and \begin{align*}M\end{align*} are complements. \begin{align*}P (N) = 96.1\%\end{align*}. Find \begin{align*}P (M)\end{align*}.
  8. \begin{align*}Q\end{align*} and \begin{align*}Z\end{align*} are complements. \begin{align*}P (Q) = \frac{1}{5}\end{align*}. Find \begin{align*}P (Z)\end{align*}.

Write complementary or not complementary.

  1. Percentage of votes that 2 candidates get in a 2-candidate election
  2. Percentage of votes that 3 candidates get in a 3-candidate election
  3. Winning a game or losing a game
  4. Choosing an odd or even number
  5. Choosing a number between 1 and 5
  6. Passing or failing a test
  7. Choosing a color of paint

Review (Answers)

To see the Review answers, open this PDF file and look for section 12.6.

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Dec 21, 2012
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