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Complement Rule for Probability

If P(A) is is the probability that event A happens then the probability that event A doesn't happen is 1-P(A).

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Finding Probability by Finding the Complement

Suppose you were asked to calculate the probability of rolling two dice and getting a different number on each. How could you find the answer without needing to enumerate all of the possibilities?

Credit: Nemo
Source: http://pixabay.com/en/red-icon-two-recreation-cartoon-25637/?oq=dice
License: CC BY-NC 3.0

This lesson is about a shortcut to the calculation of some probabilities. We’ll return to this question after the lesson.

Finding Probability Through Complements 

Sometimes the probability of an event is difficult or impossible to calculate directly. Particularly when calculating the probability of “at least one…” types of problems, or when the sample space of the complement is smaller than that of the event, it may be worth looking first for the probability of the complement of the event you are trying to measure.

Recall that the complement of an event is the sample space containing all the outcomes that are not a part of the event itself. That means that the probability of an event + the probability of the complement = 100% or 1.00, or, to say the same thing as a formula: \begin{align*}P(A)+P(A^\prime)=1\end{align*}P(A)+P(A)=1. Once you know the probability of the complement, you can just subtract it from 1 to find the probability of the event.

 

Calculating Probability 

1. What is the probability of randomly drawing a king other than the King of Hearts from a bag containing one king of each standard suit: hearts, clubs, diamonds, and spades.

Let’s say that \begin{align*}P(KH)\end{align*}P(KH) is the probability that you do draw the King of Hearts. In that case, \begin{align*}P(KH^\prime)\end{align*}P(KH) is the probability of not drawing the King of Hearts, the complement of \begin{align*}P(KH)\end{align*}P(KH). Since there are 4 kings and only 1 is the King of Hearts, we can say:

\begin{align*}P(KH)=\frac{1}{4}\end{align*}P(KH)=14

therefore:

\begin{align*}P(KH^\prime)=1-\frac{1}{4} \ or \ \frac{3}{4}=75 \%\end{align*}P(KH)=114 or 34=75%

So the probability of not drawing the King of Hearts is 75%

2. What is the probability of getting a cherry sour from a bag that starts with 9 cherry sours, 5 lemon sours, and 8 lime sours, given that you keep anything you choose and you choose up to 3 times?

 

Credit: Mariolh
Source: http://pixabay.com/en/jellies-candies-sweets-colors-204206/?oq=cAndy
License: CC BY-NC 3.0

This problem combines conditional probabilities with complements. A key point to this problem is the “keep anything you choose” part. Because we are not putting back the candy after each pull, the probability changes each time, and the chance to get a cherry improves each time we don’t choose one. Let’s deal with each pull separately at first:

  1. \begin{align*}P(C1)=\frac{9 \ cherry}{22 \ candies}=\frac{9}{22} \ or \ 41 \%\end{align*}P(C1)=9 cherry22 candies=922 or 41%First pull:
  2. \begin{align*}P(C2)=\frac{9 \ cherry}{21 \ candies}=\frac{9}{21} \ or \ 43 \%\end{align*}P(C2)=9 cherry21 candies=921 or 43%Second pull:
  3. \begin{align*}P(C3)=\frac{9 \ cherry}{20 \ candies}=\frac{9}{20} \ or \ 45 \%\end{align*}P(C3)=9 cherry20 candies=920 or 45%Third Pull:

It is tempting to think that the probability of getting at least 1 cherry is just the sum of the three probabilities, but obviously there can’t be a \begin{align*}41 \%+43 \%+45 \%=129 \%\end{align*}41%+43%+45%=129% chance. The twist is that we only choose a 2nd or 3rd time if we don’t get a cherry the time before. That means we need to calculate the chance of choosing a 2nd or 3rd time and multiply the probability of a cherry on that pull by the chance of pulling that time at all.

Fortunately we can use the complement rule to save time. The chance of needing a 2nd pull is the same as the chance that we don’t pull a cherry on the 1st pull, in other words:

\begin{align*}P(2nd \ pull)=P(C1^\prime)=1-41 \%=59 \%\end{align*}P(2nd pull)=P(C1)=141%=59%

The chance of needing a 3rd pull is the same as not getting a cherry the 2nd time:

\begin{align*}P(3rd \ pull)=P(C1^\prime) \times P(C2^\prime)=59 \% \times 57 \%=34 \% \end{align*}P(3rd pull)=P(C1)×P(C2)=59%×57%=34%

Now we can find the overall probability of getting a cherry in three pulls or less:

\begin{align*} P&(cherry) =\text{(chance of cherry of pull 1)} \\ + \text{(chance of }& \text{needing pull 2} \times \text{chance of cherry on pull 2)} \\ + \text{(chance of }& \text{needing pull 3} \times \text{chance of cherry on pull 3)} \\ P(cherry)&=41 \%+(59 \% \times 43 \%)+(34 \% \times 45 \%)=82 \%\end{align*}P+(chance of +(chance of P(cherry)(cherry)=(chance of cherry of pull 1)needing pull 2×chance of cherry on pull 2)needing pull 3×chance of cherry on pull 3)=41%+(59%×43%)+(34%×45%)=82%

3. What is the probability of rolling two dice and at least one die showing a factor of 6?

This is an “at least one…” problem. To satisfy the requirements, either one of the two dice or both need to land on 1, 2, 3, or 6. That is quite a few possibilities to solve for! However, there are many fewer possible outcomes where neither die shows 1, 2, 3, or 6 – in other words, where both dice show 4 or 5! There are only 4 ways that could happen:

  1. 1st die rolls 5 and 2nd rolls 4
  2. 1st die rolls 4 and 2nd rolls 5
  3. Both dice roll 4
  4. Both dice roll 5

There are \begin{align*}6 \times 6=36 \end{align*}6×6=36 total possible outcomes. If we say that the event “at least one die shows a factor of 6” is \begin{align*}A\end{align*}A, then \begin{align*}A^\prime\end{align*}A would be the complement, so we can say:

\begin{align*}P(A^\prime)=\frac{4 \ favorable \ outcomes}{36 \ total \ outcomes}=\frac{1}{9}\end{align*}P(A)=4 favorable outcomes36 total outcomes=19

If the complement of the event we want to calculate is \begin{align*}\frac{1}{9}\end{align*}19, then the event itself can be calculated as:

\begin{align*}P(A)=1-P(A^\prime)=\frac{8}{9} \ or \ 88.9 \%\end{align*}P(A)=1P(A)=89 or 88.9%

Therefore, we can say that the probability of rolling two dice and getting at least one factor of six is \begin{align*}\frac{8}{9}\end{align*}89.

Earlier Problem Revisited

Suppose you were asked to calculate the probability of rolling two dice and getting a different number on each. How could you find the answer without needing to enumerate all of the possibilities?

The probability of rolling two not matching numbers is the complement of rolling a matching pair. Since there are only 6 numbers to make matches from, there are only 6 matching pairs. The total number of possible outcomes of two dice is \begin{align*}6 \times 6=36\end{align*}6×6=36.

\begin{align*}\therefore P(matching)=\frac{6\text{ matches}}{36\text{ possibilities}}=\frac{1}{6} \ or \ .167\end{align*}

That means that the complement, rolling not matching numbers is:

\begin{align*}1-.167=.833 \ or \ 83.3 \%\end{align*}

That was quite a bit faster than trying to enumerate all of the possible non-matching rolls!

 

Examples 

Example 1

What is the probability of rolling a number other than 5 on a number cube?

There are six sides on a number cube, so the probability of rolling any single number is \begin{align*}\frac{1}{6}\end{align*}. Therefore, we can say: \begin{align*}P(5)=\frac{1}{6}\end{align*}  so \begin{align*}P(5^\prime)=\frac{5}{6}\end{align*} Therefore the probability of not rolling a 5 is \begin{align*}\frac{5}{6}\end{align*}.

Example 2

What is the probability of choosing a card that isn't a club from a standard deck?

There are four suits, so the probability of choosing a club is \begin{align*}\frac{1}{4}\end{align*}. The complement is the probability of not choosing a club, and it is \begin{align*}1-\frac{1}{4}=\frac{3}{4} \ or \ 75 \%\end{align*}

Example 3

What is the probability of not rolling a factor of 10 on a single roll of a 20 sided die?

There are 4 factors of 10: 1, 2, 5, and 10. Therefore the probability of rolling a factor of 10 on a 20-sided die is \begin{align*}\frac{4}{20}\end{align*}.  The complement is the probability of not rolling one of the four factors: \begin{align*}1-\frac{4}{20}=\frac{16}{20}\end{align*}

Example 4

You have a bag of crazy-flavor jellybeans. You know that the flavors are distributed as: 12 earwax, 14 belly-button lint, 26 dog food, 38 gym sock, and 10 of your favorite fruit in a the bag. What is the probability tha tyou will be unhappy with the taste of our choice if you choose one jellybean at random?

Let \begin{align*}P(F)\end{align*} be the probability of pulling a fruit-flavor, then \begin{align*}P(F^\prime)\end{align*} is the probability of not pulling a fruit flavor (and getting something disgusting instead!). There are 10 fruit-flavored beans in the bag of 100 beans, so:

\begin{align*}P(F)&=\frac{10}{100} \\ P(F^\prime)&=1-P(F)=\frac{90}{100}\end{align*}

Therefore, you have a 90% probability of being unhappy with your choice.

 

The factors of ten or twelve are: 1, 2, 3, 4, 5, 6, 10, and 12. Since all six numbers from a standard die are in that set, the probability of rolling one of them is 100% or 1.0. Therefore, the probability of not rolling one of them is \begin{align*}1-1=0\end{align*}.

Review 

  1. If you randomly pull a single card from a standard deck, what is the probability that the card is anything other than a king?
  2. What is the probability of not pulling a face card from a standard deck?
  3. What is the probability that a single roll of a 10-sided die will not land on a 7?
  4. What is the probability that a single roll of a standard die will be 1, 2, 3, 4, or 5?
  5. A candy jar contains 6 red, 7 blue, 8, green, and 9 yellow candies. What is the probability that choosing a single candy at random will result in a piece that is either red, blue, or green?
  6. What is the probability that a single choice from the jar in Q 5 will result in a piece that is either red or yellow?
  7. What is the probability that a single choice from the same jar will not be blue?
  8. You have $39 in cash, composed of the largest bills possible. What is the probability that a randomly chosen bill from the $39 will not be a $1 bill?
  9. There are 450 songs on the .mp3 player you share with your father and sister. If your dad has 125 80’s songs, and your sister has twice that many country music songs, what is the percent probability that a randomly chosen song will not be one of yours?
  10. What is the percent probability that a random roll of a fair die will not result in an even or prime number?
  11. The train station has 47 active trains. 5 are late by less than 10 minutes, 4 are between 11 and 15 minutes late, and 10 are more than 15 minutes late. What is the probability that a randomly chosen train will be on time?
  12. The probability that a student has called in sick and that it is Monday is 12%. The probability that it is Monday and not another day of the school week is 20% (there are only five days in the school week). What is the probability that a student has not called in sick, given that it is Monday?
  13. A neighborhood wanted to improve its parks so it surveyed kids to find out whether or not they rode bikes or skateboards. Out of 2300 children in the neighborhood that ride something, 1800 rode bikes, and 500 rode skateboards, while 200 of those ride both a bike and skateboard. What is the probability that a student does not ride a skateboard, given that he or she rides a bike?
  14. A movie theatre is curious about how many of its patrons buy food, how many buy a drink, and how many buy both. They track 300 people through the concessions stand one evening, out of the 300, 78 buy food only, 113 buy a drink only and the remainder buy both. What is the probability that a patron does not buy a drink if they have already bought food?

Review (Answers)

To view the Review answers, open this PDF file and look for section 6.8. 

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Vocabulary

complement

A mutually exclusive pair of events are complements to each other. For example: If the desired outcome is heads on a flipped coin, the complement is tails.

Complement rule

The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1, or for the event A, P(A) + P(A') = 1.

enumerate

Enumerate means to catalogue or list members independently.

Venn diagrams

A diagram of overlapping circles that shows the relationship among members of different sets.

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