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13.4: Probability and Permutations

Difficulty Level: At Grade Created by: CK-12
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What if you were asked to pick the final three winners in a talent contest? In how many ways could you pick a first, second, and third place winner from 20 contestants? After completing this Concept, you'll be able to calculate probabilities of events like this one using permutations.

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CK-12 Foundation: Probability and Permutations


When we talk about probability, we are talking about chance. We may use words like ‘possible’ to indicate something has a low to medium chance of happening, or ‘probable’ to indicate something has a high chance of happening.

When we perform calculations in probability we are generally looking at a situation to find the number of favorable outcomes in relation to the total number of all possible outcomes. In mathematics, a favorable outcome simply means the outcome we are looking to solve for. Over the course of 100 days a child may get ice-cream on 25 of those days. A mathematician would look at that data and conclude that the fraction of days when the child gets ice-cream is one-fourth. He may also conclude that on any day the child has a “one in four chance” of receiving ice-cream. This is one way to think about probability: the number of times something favorable happens divided by the total number of outcomes. We have no way of knowing for sure if the child is going to get ice-cream today, but we can estimate the chance that he will by looking at the number of times he did get ice-cream and the number of times he did not.

Example A

In a previous Concept, we saw an example where Nadia and Peter are going to watch two movies on a rainy Saturday. Nadia will choose the first movie, and Peter gets to choose the second. The four movies they have to choose from are The Lion King, Aladdin, Toy Story and Pinocchio. Given that Peter will choose a different movie than Nadia, what is the probability that Aladdin is the second movie they will watch?


Previously, we found that there are a total of 12 possible outcomes. Out of those 12, Aladdin was the second movie 3 times. So the probability is as follows:

\begin{align*}P(\text{Aladdin being second}) = \frac{3}{12} = \frac{1}{4} =\end{align*}P(Aladdin being second)=312=14= one in four or 25%

Example B

The card game “21 hearts” consists of dealing two cards to a player and counting the points as follows: face cards (King, Queen, Jack) are worth 10 points, and number cards are worth their face value except for aces, which are worth 11. The maximum score is 21. If the game is played only with the 13 cards in the hearts suit, what is the probability that a player will score 21?


To find the answer, we need to know two pieces of information: 1) the total number of permutations it’s possible to get in the game and 2) the number of permutations that will score 21.

To find the total number of permutations for the game, use the formula \begin{align*}_nP_r\end{align*}nPr with \begin{align*}n = 13\end{align*}n=13 and \begin{align*}r = 2\end{align*}r=2:

\begin{align*}{_{13}}P_2 = \frac{13!}{(13-2)!} =\frac{13!}{11!}= 13 \times 12 = 132 \ \text{permutations}\end{align*}13P2=13!(132)!=13!11!=13×12=132 permutations

Now we need to determine how many of those permutations score 21. We can do that by simply listing all ways to score 21. Remember, of course, that as we are talking about permutations, the combinations (ace, king) and (king, ace) each count as separate hands! The winning permutations are:

\begin{align*}& (\text{ace, king}) \qquad (\text{king, ace}) \qquad (\text{ace, queen}) \qquad (\text{queen, ace})\\ & (\text{ace, jack}) \qquad (\text{jack, ace}) \qquad \quad (\text{ace}, 10) \qquad \qquad (10, \text{ace})\end{align*}(ace, king)(king, ace)(ace, queen)(queen, ace)(ace, jack)(jack, ace)(ace,10)(10,ace)

There are 8 winning hands, so the probability of scoring 21 is given by:

\begin{align*}P(21) = \frac{8}{132} = \frac{2}{33}=\end{align*}P(21)=8132=233= two in thirty-three or approximately 6%.

Sometimes when looking at probability (especially in games of chance), there are too many wining permutations to calculate directly. In such circumstances it can be useful to remember that a player must either win or lose – it’s impossible to do both! In these circumstances remember that:

\begin{align*}P(\text{winning}) + P(\text{losing}) = 1\end{align*}P(winning)+P(losing)=1

Since it’s certain that you’ll either win or lose, the probability that you’ll win and the probability that you’ll lose add up to 1.

So to calculate the probability of losing we can use:

\begin{align*}P(\text{winning}) =1 - P(\text{losing})\end{align*}P(winning)=1P(losing)

Example C

A funfair game consists of throwing three darts at a board with 16 numbered squares in a \begin{align*}4 \times 4\end{align*}4×4 grid. The squares are numbered 1 through 16 and no number is repeated or omitted. In order to win a player needs to score 9 or more. If a dart hits a square that has already been taken or if a dart misses the board the player must throw the dart again. What are the chances of winning the game?

There are too many winning permutations to list easily, but there are only a few losing permutations. If you need a score of 9 or more to win, then you’ll lose with a score of 8 or less. There are only four combinations of numbers that add up to 8 or less (remember that you can’t repeat a number), and they can occur in any of the following orders:

\begin{align*}& \mathbf{1, 2, 3 (=6)} \qquad 1, 3, 2 \qquad 2, 1, 3 \qquad 2, 3, 1 \qquad 3, 1, 2 \qquad 3, 2, 1\\ & \mathbf{1, 2, 4 (=7)} \qquad 1, 4, 2 \qquad 2, 1, 4 \qquad 2, 4, 1 \qquad 4, 1, 2 \qquad 4, 2, 1\\ & \mathbf{1, 2, 5 (=8)} \qquad 1, 5, 2 \qquad 2, 1, 5 \qquad 2, 5, 1 \qquad 5, 1, 2 \qquad 5, 2, 1\\ & \mathbf{1, 3, 4 (=8)} \qquad 1, 4, 3 \qquad 3, 1, 4 \qquad 3, 4, 1 \qquad 4, 1, 3 \qquad 4, 3, 1\end{align*}

So there are 24 permutations totaling less than 9.

The total number of ways to choose 3 numbers out of 16 is

\begin{align*}{_{16}}P_3 = \frac{16!}{16-3!}= \frac{16!}{13!} = 16 \times 15 \times 14 = 3360 \ \text{permutations}\end{align*}

So the probability of losing the game is \begin{align*}\frac{24}{3360}\end{align*}, or \begin{align*}\frac{1}{140}\end{align*}. That means the probability of winning is:

\begin{align*}P(\text{winning}) = 1 - P(\text{losing}) = 1 - \frac{24}{3360} = 1 - \frac{1}{140} = \frac{139}{140}\end{align*} or approximately 99.3%.

Watch this video for help with the Examples above.

CK-12 Foundation: Probability and Permutations


  • The probability of a set of events is found by

\begin{align*}\text{Probability} = \frac{\text{number of matching events in sample place}}{\text{number of total events in sample place}}\end{align*}

  • A permutation is when we are choosing \begin{align*}r\end{align*} ordered items from a group of \begin{align*}n\end{align*} items, where the order chosen matters. The number of permutations is given by the formula:

\begin{align*}{_n}P_r = \frac{n!}{(n-r)!}\end{align*}

Guided Practice

What is the probability that a randomly generated 3-letter arrangement of the letters \begin{align*}M\end{align*}, \begin{align*}M\end{align*}, and \begin{align*}O\end{align*} will result in spelling the word \begin{align*}MOM\end{align*}?


To find the probability, we first must describe the sample space. It might be tempting to think that the only possibilities are \begin{align*}MMO\end{align*}, \begin{align*}MOM\end{align*} and \begin{align*}OMM\end{align*}. However, we must treat each \begin{align*}M\end{align*} as an distinct letter. Let's think of them as \begin{align*}M_1\end{align*} and \begin{align*}M_2\end{align*}. Since we have 3 distinct objects and we want to order all three of them, this is:

\begin{align*}_3P_3=3!=3\cdot 2 \cdot 1=6\end{align*}

Thus, there are 6 ways to order the 3 letters. Let's see what that looks like:

\begin{align*} M_1 M_2 O \\ MOM=M_1 O M_2\\ M_2 M_1 O \\ MOM= M_2 O M_1\\ O M_1 M_2\\ O M_2 M_1\end{align*}

2 of these 6 orderings spell \begin{align*}MOM\end{align*}, which means the probability of spelling \begin{align*}MOM\end{align*} is:

\begin{align*}P(MOM)=\frac{2}{6}\approx 0.67\end{align*}.


For 1-5, find the number of permutations.

  1. \begin{align*}_5P_2\end{align*}
  2. \begin{align*}_9P_4\end{align*}
  3. \begin{align*}_{11}P_5\end{align*}
  4. How many ways can you plant a rose bush, a lavender bush and a hydrangea bush in a row?
  5. How many ways can you pick a president, a vice president, a secretary and a treasurer out of 28 people for student council?

For 6-10, find the probabilities.

  1. What is the probability that a randomly generated 3-letter arrangement of the letters A,E,L, Q and U will result in spelling the word EQUAL?
  2. What is the probability that a randomly generated 3-letter arrangement of the letters in the word SPIN ends with the letter N?
  3. A bag contains eight chips numbered 1 through 8. Two chips are drawn randomly from the bag and laid down in the order they were drawn. What is the probability that the 2-digit number formed is divisible by 3?
  4. A prepaid telephone calling card comes with a randomly selected 4-digit PIN, using the digits 1 through 9 without repeating any digits. What is the probability that the PIN for a card chosen at random does not contain the number 7?
  5. Janine makes a playlist of 8 songs and has her computer randomly shuffle them. If one song is by Little Bow Wow, what is the probability that this song will play first?

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combination Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.
Favorable Outcome A favorable outcome is the outcome that you are looking for in an experiment.
Permutation A permutation is an arrangement of objects where order is important.
Probability Probability is the chance that something will happen. It can be written as a fraction, decimal or percent.

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Difficulty Level:
At Grade
Date Created:
Oct 01, 2012
Last Modified:
Apr 11, 2016
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