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The Binomial Distribution

Identify binomial experiments.

Learning Objectives

After completing this section, you should be able to:

Identify binomial experiments.
Use the binomial distribution to analyze binomial experiments.

Binomial Experiments

The tools of this section apply to multistage experiments that satisfy some pretty specific criteria. Before we move on to the analysis, we need to introduce and explain those criteria so that we can recognize experiments that fall into this category. Experiments that satisfy each of these criteria are called binomial experiments. A binomial experiment is an experiment with a fixed number of repeated independent binomial trials, where each trial has the same probability of success.

The first criterion involves the structure of the stages. Each stage of the experiment should be a replication of every other stage; we call these replications trials. An example of this is flipping a coin 10 times; each of the ten flips is a trial, and they all occur under the same conditions as every other. Further, each trial must have only two possible outcomes. These two outcomes are typically labeled “success” and “failure,” even if there is not a positive or negative connotation associated with those outcomes. Experiments with more than two outcomes in their sample spaces are sometimes reconsidered in a way that forces just two outcomes; all we need to do is completely divide the sample space into two parts that we can label “success” and “failure.” For example, your grade on an exam might be recorded as A, B, C, D, or F, but we could instead think of the grades A, B, C, and D as “success” and a grade of F as “failure.” Trials with only two outcomes are called binomial trials (the word binomial derives from Latin and Greek roots that mean “two parts”).

The next criterion that we’ll be looking for is independence of trials. Back in Tree Diagrams, Tables, and Outcomes, we said that two stages of an experiment are independent if the outcome of one stage doesn’t affect the other stage. Independence is necessary for the experiments we want to analyze in this section.

Next, we require that the number of trials in the experiment be decided before the experiment begins. For example, we might say “flip a coin 10 times.” The number of trials there is fixed at 10. However, if we say “flip a coin until you get 5 heads,” then the number of trials could be as low as 5, but theoretically it could be 50 or a 100 (or more)! We can’t apply the tools from this section in cases where the number of trials is indeterminate.

Condensed — the full section is in OpenStax Contemporary Mathematics.

The Binomial Formula

If we flip a coin 100 times, you might expect the number of heads to be around 50, but you probably wouldn’t be surprised if the actual number of heads was 47 or 52. What is the probability that the number of heads is exactly 50? Or falls between 45 and 55? It seems unlikely that we would get more than 70 heads. Exactly how unlikely is that?

Each of these questions is a question about the number of successes in a binomial experiment (flip a coin 100 times, “success” is flipping heads). We could theoretically use the techniques we’ve seen in earlier sections to answer each of these, but the number of calculations we’d have to do is astronomical; just building the tree diagram that represents this situation is more than we could complete in a lifetime; it would have \({2}^{100}\approx 1.3\times {10}^{30}\) final nodes! To put that number in perspective, if we could draw 1,000 dots every second, and we started at the moment of the Big Bang, we’d currently be about 0.00000003% of the way to drawing out those final nodes. Luckily, there’s a shortcut called the Binomial Formula that allows us to get around doing all those calculations!

We can use this formula to answer one of our questions about 100 coin flips. What is the probability of flipping exactly 50 heads? In this case, \(n=100\), \(p=\frac{1}{2}\), and \(a=50\), so \(P(\text{flip 50 heads})=C100{}_{50}\times {(\frac{1}{2})}^{50}\times {(1-\frac{1}{2})}^{100-50}\). Unfortunately, many calculators will balk at this calculation; that first factor (\({}_{100}{C}_{50}\)) is an enormous number, and the other two factors are very close to zero. Even if your calculator can handle numbers that large or small, the arithmetic can create serious errors in rounding off.

Condensed — the full section is in OpenStax Contemporary Mathematics.

The Binomial Distribution

If we are interested in the probability of more than just a single outcome in a binomial experiment, it’s helpful to think of the Binomial Formula as a function, whose input is the number of successes and whose output is the probability of observing that many successes. Generally, for a small number of trials, we’ll give that function in table form, with a complete list of the possible outcomes in one column and the probability in the other.

For example, suppose Kristen is practicing her basketball free throws. Assume Kristen always makes 82% of those shots. If she attempts 5 free throws, then the Binomial Formula gives us these probabilities:

Shots Made	Probability
0	0.000189
1	0.004304
2	0.0392144
3	0.1786432
4	0.4069096
5	0.3707398

A table that lists all possible outcomes of an experiment along with the probabilities of those outcomes is an example of a probability density function (PDF). A PDF may also be a formula that you can use to find the probability of any outcome of an experiment.

If we want to know the probability of a range of outcomes, we could add up the corresponding probabilities. Going back to Kristen’s free throws, we can find the probability that she makes 3 or fewer of her 5 attempts by adding up the probabilities associated with the corresponding outcomes (in this case: 0, 1, 2, or 3):

\[\begin{array}{lll}P(\text{makes 3 or fewer}) & = & P(a=0)+P(a=1)+P(a=2)+P(a=3) \\ & = & 0.000189+0.004304+0.0392144+0.1786432 \\ & = & 0.2223506\end{array}\]

The probability that the outcome of an experiment is less than or equal to a given number is called a cumulative probability. A table of the cumulative probabilities of all possible outcomes of an experiment is an example of a cumulative distribution function (CDF). A CDF may also be a formula that you can use to find those cumulative probabilities.

Here are the PDF and CDF for Kristen’s free throws:

Shots Made	Probability	Cumulative
0	0.000189	0.000189
1	0.004304	0.004493
2	0.0392144	0.0437073
3	0.1786432	0.2223506
4	0.4069096	0.6292602
5	0.3707398	1

Condensed — the full section is in OpenStax Contemporary Mathematics.

Key Concepts

Binomial experiments result when we count the number of successful outcomes in a fixed number of repeated, independent trials with a constant probability of success.
The binomial distribution is used to find probabilities associated with binomial experiments.
Probability density functions (PDFs) describe the probabilities of individual outcomes in an experiment; cumulative distribution functions (CDFs) give the probabilities of ranges of outcomes.

Formulas

Suppose we have a binomial experiment with \(n\) trials and the probability of success in each trial is \(p\). Then:
\[P(\text{number of successes is}\ a)=Cn{}_{a}\times {p}^{a}\times {(1-p)}^{n-a}\]

Practice (3)

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

Decide whether each of the following is a binomial experiment. For those that aren’t, identify which criterion or criteria are violated.
1. You roll a standard 6-sided die 10 times and write down the number that appears each time.
2. You roll a standard 6-sided die 10 times and write down whether the die shows a 6 or not.
3. You roll a standard 6-sided die until you get a 6.
4. You roll a standard 6-sided die 10 times. On the first roll, we define “success” as rolling a 4 or greater. After the first roll, we define “success” as rolling a number greater than the result of the previous roll.
જવાબ બતાવો
1. Since we’re noting 1 of 6 possible outcomes, the trials are not binomial. So, this isn’t a binomial experiment.
2. We have 2 possible outcomes (“6” and “not 6”), the trials are independent, the probability of success is the same every time, and the number of trials is fixed. This is a binomial experiment.
3. Since the number of trials isn’t fixed (we don’t know if we’ll get our first 6 after 1 roll or 20 rolls or somewhere in between), this isn’t a binomial experiment.
4. Here, the probability of success might change with every roll (on the first roll, that probability is \(\frac{1}{2}\); if the first roll is a 6, the probability of success on the next roll is zero). So, this is not a binomial experiment.
1. Find the probability of rolling a standard 6-sided die 4 times and getting exactly one 6 without using technology.
2. Find the probability of rolling a standard 6-sided die 60 times and getting exactly ten 6s using technology.
3. Find the probability of rolling a standard 6-sided die 60 times and getting exactly eight 6s using technology.
જવાબ બતાવો
1. We’ll apply the Binomial Formula, where \(n=4\), \(a=1\), and \(p=\frac{1}{6}\): \[\begin{array}{lll}P(\text{rolling one 6}) & = & {C}_{1}4\times {(\frac{1}{6})}^{1}\times {(\frac{5}{6})}^{4-1} \\ & = & \frac{4!}{1!(4-1)!}\times \frac{1}{6}\times {(\frac{5}{6})}^{3} \\ & = & 4\times \frac{1}{6}\times \frac{{5}^{3}}{{6}^{3}} \\ & = & \frac{4\times {5}^{3}}{{6}^{4}} \\ & = & \frac{500}{1,296}.\end{array}\]
2. Here, \(n=60\), \(a=10\), and \(p=\frac{1}{6}\). In Google Sheets, we’ll enter “=BINOMDIST(10, 60, 1/6, FALSE)” to get our result: 0.137.
3. This experiment is the same as in Exercise 2 of this example; we’re simply changing the number of successes from 10 to 8. Making that change in the formula in Google Sheets, we get the probability 0.116.
Suppose we are about to flip a fair coin 50 times. Let \(H\) represent the number of heads that result from those flips. Use technology to find the following:
1. \(P(H\le 22)\)
2. \(P(H<26)\)
3. \(P(H>28)\)
4. \(P(H\ge 20)\)
5. \(P(20
જવાબ બતાવો
1. The event here is defined by \(H\le 22\), which is the inequality we need to have if we want to use the Binomial CDF. In Google Sheets, we’ll enter “=BINOMDIST(22, 50, 0.5, TRUE)” to get our answer: 0.2399.
2. This event uses the wrong inequality, so we need to do some preliminary work. If \(H<26\), that means \(H\le 25\) (because \(H\) has to be a whole number). So, we’ll enter “=BINOMDIST(25, 50, 0.5, TRUE)” to find \(P(H<26)=P(H\le 25)=0.5561\).
3. The inequality associated with this event is pointing in the wrong direction. If \(E\) is the event \(H>28\), that means that \(E\) contains the outcomes {29, 30, 31, 32, 33, …}. Thus, \(E'\) must contain the outcomes {…, 25, 26, 27, 28}. In other words, \(E'\) is defined by \(H\le 28\). Since it uses \(\le\), we can find \(P(E')\) using “=BINOMDIST(28, 50, 0.5, TRUE)”: 0.8389 So, using the formula for probabilities of complements, we have \[\begin{array}{l}P(E)=1-P(E') \\ =1-0.8389 \\ =0.1611.\end{array}\]
4. As in part 3, this inequality is pointing in the wrong direction. If \(F\) is the event \(H\ge 20\), then \(F\) contains the outcomes {20, 21, 22, 23, …}. That means \(F'\) contains the outcomes {…, 16, 17, 18, 19}, and so \(F'\) is defined by \(H\le 19\). So, we can find \(P(F')\) using “=BINOMDIST(19, 50, 0.5, TRUE)”: 0.0595. Finally, using the formula for probabilities of complements, we get: \[\begin{array}{l}P(F)=1-P(F') \\ =1-0.0595 \\ =0.9405.\end{array}\]
5. If \(20

Symbols used here

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

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

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\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.

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: The Binomial Distribution

Identify binomial experiments.
Use the binomial distribution to analyze binomial experiments.
You roll a standard 6-sided die 10 times and write down the number that appears each time.
You roll a standard 6-sided die 10 times and write down whether the die shows a 6 or not.
You roll a standard 6-sided die until you get a 6.
You roll a standard 6-sided die 10 times. On the first roll, we define “success” as rolling a 4 or greater. After the first roll, we define “success” as rolling a number greater than the result of the previous roll.
Since we’re noting 1 of 6 possible outcomes, the trials are not binomial. So, this isn’t a binomial experiment.
We have 2 possible outcomes (“6” and “not 6”), the trials are independent, the probability of success is the same every time, and the number of trials is fixed. This is a binomial experiment.

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.

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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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