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Probability with Permutations and Combinations

Calculate probabilities with permutations.

Learning Objectives

After completing this section, you should be able to:

Calculate probabilities with permutations.
Calculate probabilities with combinations.

Using Permutations to Compute Probabilities

Recall that we can use permutations to count how many ways there are to put a number of items from a list in order. If we’re looking at an experiment whose sample space looks like an ordered list, then permutations can help us to find the right probabilities.

Using Permutations to Compute Probabilities

Try it.

In horse racing, an exacta bet is one where the player tries to predict the top two finishers in particular race in order. If there are 9 horses in a race, and a player decided to make an exacta bet at random, what is the probability that they win?
You are in a club with 10 people, 3 of whom are close friends of yours. If the officers of this club are chosen at random, what is the probability that you are named president and one of your friends is named vice president?
A bag contains slips of paper with letters written on them as follows: A, A, B, B, B, C, C, D, D, D, D, E. If you draw 3 slips, what is the probability that the letters will spell out (in order) the word BAD?

Solution

Since order matters for this situation, we’ll use permutations. How many different exacta bets can be made? Since there are 9 horses and we must select 2 in order, we know there are ${}_{9}{P}_{2}=72$ possible outcomes. That’s the size of our sample space, so it will go in the denominator of the probability. Since only one of those outcomes is a winner, the numerator of the probability is 1. So, the probability of randomly selecting the winning exacta bet is $\frac{1}{72}$.
There are 10 people in the club, and 2 will be chosen to be officers. Since the order matters, there are ${}_{10}{P}_{2}=90$ different ways to select officers. Next, we must figure out how many outcomes are in our event. We’ll use the Multiplication Rule for Counting to find that number. There is only 1 choice for president in our event, and there are 3 choices for vice president. So, there are $1\times 3=3$ outcomes in the event. Thus, the probability that you will serve as president with one of your friends as vice president is $\frac{3}{90}=\frac{1}{30}$.
There are 12 slips of paper in the bag, and 3 will be drawn. So, there are ${}_{12}{P}_{3}=1320$ possible outcomes. Now, we’ll compute the number of outcomes in our event. The first letter drawn must be a B, and there are 3 of those. Next must come an A (2 of those) and then a D (4 of those). Thus, there are $3\times 2\times 4=24$ outcomes in our event. So, the probability that the letters drawn spell out the word BAD is $\frac{24}{1320}=\frac{1}{55}$.

Combinations to Computer Probabilities

If the sample space of our experiment is one in which order doesn’t matter, then we can use combinations to find the number of outcomes in that sample space.

Using Combinations to Compute Probabilities

Try it.

Palmetto Cash 5 is a game offered by the South Carolina Education Lottery. Players choose 5 numbers from the whole numbers between 1 and 38 (inclusive); the player wins the jackpot of $100,000 if the randomizer selects those numbers in any order. If you buy one ticket for this game, what is the probability that you win the top prize by choosing all 5 winning numbers?
There’s a second prize in the Palmetto Cash 5 game that a player wins if 4 of the player's 5 numbers are among the 5 winning numbers. What’s the probability of winning the second prize?
Scrabble is a word-building board game. Players make hands of 7 letters by selecting tiles with single letters printed on them blindly from a bag (2 tiles have nothing printed on them; these blanks can stand for any letter). Players use the letters in their hands to spell out words on the board. Initially, there are 100 tiles in the bag. Of those, 44 are (or could be) vowels (9 As, 12 Es, 9 Is, 8 Os, 4 Us, and 2 blanks; we’ll treat Y as a consonant). What is the probability that your initial hand has no vowels?

Solution

There are 38 numbers to choose from, and the order of the 5 we pick doesn’t matter. So, there are ${}_{38}{C}_{5}=501,492$ outcomes in the sample space. Only one outcome is in our winning event, so the probability of winning is $\frac{1}{501,492}$.
As in part 1 of this example,, there are 501,492 outcomes in the sample space. The tricky part here is figuring out how many outcomes are in our event. To qualify, the outcome must contain 4 of the 5 winning numbers, plus one losing number. There are ${}_{5}{C}_{4}=5$ ways to choose the 4 winning numbers, and there are $38-5=33$ losing numbers. So, using the Multiplication Rule for Counting, there are $5\times 33=165$ outcomes in our event. Thus, the probability of winning the second prize is $\frac{165}{501,492}=\frac{55}{167,164}$, which is about 0.00033.
The number of possible starting hands is ${}_{100}{C}_{7}=16,007,560,800$. There are $100-44=56$ consonants in the bag, so the number of all-consonant hands is ${}_{56}{C}_{7}=231,917,400$. Thus, the probability of drawing all consonants is $\frac{231,917,40}{16,007,560,800}=\frac{32,139}{2,425,388}\approx 0.0145$.

Key Concepts

We use permutations and combinations to count the number of equally likely outcomes in an event and in a sample space, which allows us to compute theoretical probabilities.

Practice (2)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. In horse racing, an exacta bet is one where the player tries to predict the top two finishers in particular race in order. If there are 9 horses in a race, and a player decided to make an exacta bet at random, what is the probability that they win?
2. You are in a club with 10 people, 3 of whom are close friends of yours. If the officers of this club are chosen at random, what is the probability that you are named president and one of your friends is named vice president?
3. A bag contains slips of paper with letters written on them as follows: A, A, B, B, B, C, C, D, D, D, D, E. If you draw 3 slips, what is the probability that the letters will spell out (in order) the word BAD?
Mutasd meg a választ!
1. Since order matters for this situation, we’ll use permutations. How many different exacta bets can be made? Since there are 9 horses and we must select 2 in order, we know there are ${}_{9}{P}_{2}=72$ possible outcomes. That’s the size of our sample space, so it will go in the denominator of the probability. Since only one of those outcomes is a winner, the numerator of the probability is 1. So, the probability of randomly selecting the winning exacta bet is $\frac{1}{72}$.
2. There are 10 people in the club, and 2 will be chosen to be officers. Since the order matters, there are ${}_{10}{P}_{2}=90$ different ways to select officers. Next, we must figure out how many outcomes are in our event. We’ll use the Multiplication Rule for Counting to find that number. There is only 1 choice for president in our event, and there are 3 choices for vice president. So, there are $1\times 3=3$ outcomes in the event. Thus, the probability that you will serve as president with one of your friends as vice president is $\frac{3}{90}=\frac{1}{30}$.
3. There are 12 slips of paper in the bag, and 3 will be drawn. So, there are ${}_{12}{P}_{3}=1320$ possible outcomes. Now, we’ll compute the number of outcomes in our event. The first letter drawn must be a B, and there are 3 of those. Next must come an A (2 of those) and then a D (4 of those). Thus, there are $3\times 2\times 4=24$ outcomes in our event. So, the probability that the letters drawn spell out the word BAD is $\frac{24}{1320}=\frac{1}{55}$.
1. Palmetto Cash 5 is a game offered by the South Carolina Education Lottery. Players choose 5 numbers from the whole numbers between 1 and 38 (inclusive); the player wins the jackpot of $100,000 if the randomizer selects those numbers in any order. If you buy one ticket for this game, what is the probability that you win the top prize by choosing all 5 winning numbers?
2. There’s a second prize in the Palmetto Cash 5 game that a player wins if 4 of the player's 5 numbers are among the 5 winning numbers. What’s the probability of winning the second prize?
3. Scrabble is a word-building board game. Players make hands of 7 letters by selecting tiles with single letters printed on them blindly from a bag (2 tiles have nothing printed on them; these blanks can stand for any letter). Players use the letters in their hands to spell out words on the board. Initially, there are 100 tiles in the bag. Of those, 44 are (or could be) vowels (9 As, 12 Es, 9 Is, 8 Os, 4 Us, and 2 blanks; we’ll treat Y as a consonant). What is the probability that your initial hand has no vowels?
Mutasd meg a választ!
1. There are 38 numbers to choose from, and the order of the 5 we pick doesn’t matter. So, there are ${}_{38}{C}_{5}=501,492$ outcomes in the sample space. Only one outcome is in our winning event, so the probability of winning is $\frac{1}{501,492}$.
2. As in part 1 of this example,, there are 501,492 outcomes in the sample space. The tricky part here is figuring out how many outcomes are in our event. To qualify, the outcome must contain 4 of the 5 winning numbers, plus one losing number. There are ${}_{5}{C}_{4}=5$ ways to choose the 4 winning numbers, and there are $38-5=33$ losing numbers. So, using the Multiplication Rule for Counting, there are $5\times 33=165$ outcomes in our event. Thus, the probability of winning the second prize is $\frac{165}{501,492}=\frac{55}{167,164}$, which is about 0.00033.
3. The number of possible starting hands is ${}_{100}{C}_{7}=16,007,560,800$. There are $100-44=56$ consonants in the bag, so the number of all-consonant hands is ${}_{56}{C}_{7}=231,917,400$. Thus, the probability of drawing all consonants is $\frac{231,917,40}{16,007,560,800}=\frac{32,139}{2,425,388}\approx 0.0145$.

Symbols used here

\approx

approximately equal
Equal to the precision shown, not exactly.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\bar{x},\ \mu

sample mean, population mean
Average of the data; average of the whole population.

\sigma,\ s,\ \sigma^2

standard deviation, sample s.d., variance
Typical distance from the mean; its square.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

E[X],\ \operatorname{Var}(X)

expected value, variance
Probability-weighted average of X; its spread.

N(\mu, \sigma^2),\ z

normal distribution, z-score
The bell curve with mean μ and variance σ²; (x − μ)/σ.

How to: Probability with Permutations and Combinations

Calculate probabilities with permutations.
Calculate probabilities with combinations.
In horse racing, an exacta bet is one where the player tries to predict the top two finishers in particular race in order. If there are 9 horses in a race, and a player decided to make an exacta bet at random, what is the probability that they win?
You are in a club with 10 people, 3 of whom are close friends of yours. If the officers of this club are chosen at random, what is the probability that you are named president and one of your friends is named vice president?
A bag contains slips of paper with letters written on them as follows: A, A, B, B, B, C, C, D, D, D, D, E. If you draw 3 slips, what is the probability that the letters will spell out (in order) the word BAD?
Since order matters for this situation, we’ll use permutations. How many different exacta bets can be made? Since there are 9 horses and we must select 2 in order, we know there are
There are 10 people in the club, and 2 will be chosen to be officers. Since the order matters, there are
There are 12 slips of paper in the bag, and 3 will be drawn. So, there are

Questions people ask

Mean or median — which should I use?

Median when the data have outliers or a long tail (incomes, house prices); mean when the data are roughly symmetric and you want every value to count. Report both if they disagree — the gap is itself information.

What does a p-value actually say?

The probability of seeing data at least this extreme if the null hypothesis were true. It is not the probability that the null hypothesis is true.

Why divide by n − 1 for the sample variance?

The sample mean sits closer to the sample than the true mean does, so squared deviations from it are slightly too small on average; dividing by n − 1 instead of n corrects the bias.

Próbáld a sajátodat.

Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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