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Binomial Distribution

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The binomial distribution is, in essence, the probability distribution of the number of heads resulting from flipping a weighted coin multiple times. It is useful for analyzing the results of repeated independent trials, especially the probability of meeting a particular threshold given a specific error rate, and thus has applications to risk management . For this reason, the binomial distribution is also important in determining statistical significance .

Formal Definition

Finding the binomial distribution, properties of the binomial distribution, practical applications, binomial test.

A Bernoulli trial , or Bernoulli experiment , is an experiment satisfying two key properties:

• There are exactly two complementary outcomes, success and failure.
• The probability of success is the same every time the experiment is repeated.

A binomial experiment is a series of \(n\) Bernoulli trials, whose outcomes are independent of each other. A random variable , \(X\), is defined as the number of successes in a binomial experiment. Finally, a binomial distribution is the probability distribution of \(X\).

For example, consider a fair coin. Flipping the coin once is a Bernoulli trial, since there are exactly two complementary outcomes (flipping a head and flipping a tail), and they are both \(\frac{1}{2}\) no matter how many times the coin is flipped. Note that the fact that the coin is fair is not necessary; flipping a weighted coin is still a Bernoulli trial.

A binomial experiment might consist of flipping the coin 100 times, with the resulting number of heads being represented by the random variable \(X\). The binomial distribution of this experiment is the probability distribution of \(X.\)

Determining the binomial distribution is straightforward but computationally tedious. If there are \(n\) Bernoulli trials, and each trial has a probability \(p\) of success, then the probability of exactly \(k\) successes is

\[\binom{n}{k}p^k(1-p)^{n-k}.\]

This is written as \(\text{Pr}(X=k)\), denoting the probability that the random variable \(X\) is equal to \(k\), or as \(b(k;n,p)\), denoting the binomial distribution with parameters \(n\) and \(p\).

The above formula is derived from choosing exactly \(k\) of the \(n\) trials to result in successes, for which there are \(\binom{n}{k}\) choices, then accounting for the fact that each of the trials marked for success has a probability \(p\) of resulting in success, and each of the trials marked for failure has a probability \(1-p\) of resulting in failure. The binomial coefficient \(\binom{n}{k}\) lends its name to the binomial distribution.

Consider a weighted coin that flips heads with probability \(0.25\). If the coin is flipped 5 times, what is the resulting binomial distribution? This binomial experiment consists of 5 trials, a \(p\)-value of \(0.25\), and the number of successes is either 0, 1, 2, 3, 4, or 5. Therefore, the above formula applies directly: \[\begin{align} \text{Pr}(X=0) &= b(0;5,0.25) = \binom{5}{0}(0.25)^0(0.75)^5 \approx 0.237\\ \text{Pr}(X=1) &= b(1;5,0.25) = \binom{5}{1}(0.25)^1(0.75)^4 \approx 0.396\\ \text{Pr}(X=2) &= b(2;5,0.25) = \binom{5}{2}(0.25)^2(0.75)^3 \approx 0.263\\ \text{Pr}(X=3) &= b(3;5,0.25) = \binom{5}{3}(0.25)^3(0.75)^2 \approx 0.088\\ \text{Pr}(X=4) &= b(4;5,0.25) = \binom{5}{4}(0.25)^4(0.75)^1 \approx 0.015\\ \text{Pr}(X=5) &= b(5;5,0.25) = \binom{5}{5}(0.25)^5(0.75)^0 \approx 0.001. \end{align}\] It's worth noting that the most likely result is to flip one head, which is explored further below when discussing the mode of the distribution. \(_\square\)

This can be represented pictorially, as in the following table:

The binomial distribution \(b(5,0.25)\)

You have an (extremely) biased coin that shows heads with probability 99% and tails with probability 1%. To test the coin, you tossed it 100 times.

What is the approximate probability that heads showed up exactly \( 99 \) times?

A fair coin is flipped 10 times. What is the probability that it lands on heads the same number of times that it lands on tails?

Give your answer to three decimal places.

There are several important values that give information about a particular probability distribution. The most important are as follows:

• The mean , or expected value , of a distribution gives useful information about what average one would expect from a large number of repeated trials.
• The median of a distribution is another measure of central tendency, useful when the distribution contains outliers (i.e. particularly large/small values) that make the mean misleading.
• The mode of a distribution is the value that has the highest probability of occurring.
• The variance of a distribution measures how "spread out" the data is. Related is the standard deviation , the square root of the variance, useful due to being in the same units as the data.

Three of these values--the mean, mode, and variance--are generally calculable for a binomial distribution. The median, however, is not generally determined.

The mean of a binomial distribution is intuitive:

The mean of \(b(n,p)\) is \(np.\)

In other words, if an unfair coin that flips heads with probability \(p\) is flipped \(n\) times, the expected result would be \(np\) heads.

Let \(X_1, X_2, \ldots, X_n\) be random variables representing the Bernoulli trial with probability \(p\) of success. Then \(X = X_1 + X_2 + \cdots + X_n\), by definition. By linearity of expectation , \[E[X]=E[X_1+X_2+\cdots+X_n]=E[X_1]+E[X_2]+\cdots+E[X_n]=\underbrace{p+p+\cdots+p}_{n\text{ times}}=np.\ _\square\]

You have an (extremely) biased coin that shows heads with 99% probability and tails with 1% probability.

If you toss it 100 times, what is the expected number of times heads will come up?

This problem is part of the set Extremely Biased Coins.

A similar strategy can be used to determine the variance:

The variance of \(b(n,p)\) is \(np(1-p)\).

Since variance is additive, a similar proof to the above can be used: \[ \begin{align*} \text{Var}[X] &= \text{Var}(X_1 + X_2 + \cdots + X_n) \\ &= \text{Var}(X_1) + \text{Var}(X_2) + \cdots + \text{Var}(X_n) \\ &= \underbrace{p(1-p)+p(1-p)+\cdots+p(1-p)}_{n\text{ times}} \\ &= np(1-p) \end{align*} \] since the variance of a single Bernoulli trial is \(p(1-p)\). \(_\square\)

The mode, however, is slightly more complicated. In most cases the mode is \(\lfloor (n+1)p \rfloor\), but if \((n+1)p\) is an integer, both \((n+1)p\) and \((n+1)p-1\) are modes. Additionally, in the trivial cases of \(p=0\) and \(p=1\), the modes are 0 and \(n,\) respectively.

The mode of \(b(n,p)\) is

\[ \text{mode} = \begin{cases} 0 & \text{if } p = 0 \\ n & \text{if } p = 1 \\ (n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\mathbb{Z} \\ \big\lfloor (n+1)\,p\big\rfloor & \text{if }(n+1)p\text{ is 0 or a non-integer}. \end{cases} \]

Daniel has a weighted coin that flips heads \(\frac{2}{5}\) of the time and tails \(\frac{3}{5}\) of the time. If he flips it \(9\) times, the probability that it will show heads exactly \(n\) times is greater than or equal to the probability that it will show heads exactly \(k\) times, for all \(k=0, 1,\dots, 9, k\ne n\).

If the probability that the coin will show heads exactly \(n\) times in \(9\) flips is \(\frac{p}{q}\) for positive coprime integers \(p\) and \(q\), then find the last three digits of \(p\).

The binomial distribution is applicable to most situations in which a specific target result is known, by designating the target as "success" and anything other than the target as "failure." Here is an example:

A die is rolled 3 times. What is the probability that no sixes occur? In this binomial experiment, rolling anything other than a 6 is a success and rolling a 6 is failure. Since there are three trials, the desired probability is \[b\left(3;3,\frac{5}{6}\right)=\binom{3}{3}\left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right)^0 \approx .579.\] This could also be done by designating rolling a 6 as a success, and rolling anything else as failure. Then the desired probability would be \[b\left(0;3,\frac{1}{6}\right)=\binom{3}{0}\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^3 \approx .579\] just as before. \(_\square\)

The binomial distribution is also useful in analyzing a range of potential results, rather than just the probability of a specific one:

A manufacturer of widgets knows that 20% of the widgets he produces are defective. If he produces 10 widgets per day, what is the probability that at most two of them are defective? In this binomial experiment, manufacturing a working widget is a success and manufacturing a defective widget is a failure. The manufacturer needs at least 8 successes, making the probability \[ \begin{align*} b(8;10,0.8)+b(9;10,0.8)+b(10;10,0.8) &=\binom{10}{8}(0.8)^8(0.2)^2+\binom{10}{9}(0.8)^9(0.2)^1+\binom{10}{10}(0.8)^{10} \\\\ &\approx 0.678. \ _\square \end{align*} \]

This example also illustrates an important clash with intuition: generally, one would expect that an 80% success rate is appropriate when requiring 8 of 10 widgets to not be defective. However, the above calculation shows that an 80% success rate only results in at least 8 successes less than 68% of the time!

This calculation is especially important for a related reason: since the manufacturer knows his error rate and his quota, he can use the binomial distribution to determine how many widgets he must produce in order to earn a sufficiently high probability of meeting his quota of non-defective widgets.

Related to the final note of the last section, the binomial test is a method of testing for statistical significance . Most commonly, it is used to reject the null hypothesis of uniformity; for example, it can be used to show that a coin or die is unfair. In other words, it is used to show that the given data is unlikely under the assumption of fairness, so that the assumption is likely false.

A coin is flipped 100 times, and the results are 61 heads and 39 tails. Is the coin fair? The null hypothesis is that the coin is fair, in which case the probability of flipping at least 61 heads is \[\sum_{i=61}^{100}b(i;100,0.5) = \sum_{i=61}^{100}\binom{100}{i}(0.5)^{100} \approx 0.0176,\] or \(1.76\%\). Determining whether this result is statistically significant depends on the desired confidence level; this would be enough to reject the null hypothesis at the 5% level, but not the 1% one. As the most commonly used confidence level is the 5% one, this would generally be considered sufficient to conclude that the coin is unfair. \(_\square\)

• Geometric Distribution
• Poisson Distribution

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Binomial distribution

by Marco Taboga , PhD

The binomial distribution is a univariate discrete distribution used to model the number of favorable outcomes obtained in a repeated experiment.

Table of contents

How the distribution is used

What you need to know, relation to the bernoulli distribution, expected value, moment generating function, characteristic function, distribution function, solved exercises.

Consider an experiment having two possible outcomes: either success or failure.

Suppose that the experiment is repeated several times and the repetitions are independent of each other.

The distribution of the number of experiments in which the outcome turns out to be a success is called binomial distribution.

Chart of binomial distribution with interactive calculator

A binomial distribution can be seen as a sum of mutually independent Bernoulli random variables that take value 1 in case of success of the experiment and value 0 otherwise.

This connection between the binomial and Bernoulli distributions will be illustrated in detail in the remainder of this lecture and will be used to prove several properties of the binomial distribution.

Before proceeding, you are advised to study the lecture on the Bernoulli distribution .

The binomial distribution is characterized as follows.

The binomial distribution is intimately related to the Bernoulli distribution. The following propositions show how.

binocdf(x,n,p)

returns the value of the distribution function at the point x when the parameters of the distribution are n and p .

You can also use the calculator at the top of this page.

Below you can find some exercises with explained solutions.

How to cite

Please cite as:

Taboga, Marco (2021). "Binomial distribution", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix. https://www.statlect.com/probability-distributions/binomial-distribution.

Most of the learning materials found on this website are now available in a traditional textbook format.

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Statistics By Jim

Making statistics intuitive

Binomial Distribution: Uses & Calculator

By Jim Frost 2 Comments

What is the Binomial Distribution?

The binomial distribution is a discrete probability distribution that calculates the likelihood an event will occur a specific number of times in a set number of opportunities. Use this distribution when you have a binomial random variable. These variables count how often an event occurs within a fixed number of trials. They have only two possible outcomes that are mutually exclusive.

For example, the binomial probability distribution can answer the following questions. What is the probability of getting:

• Six heads when you toss the coin ten times?
• 12 women in a sample size of 20?
• Three defective items in a batch of 100?
• Two flu infections over 20 years?

This distribution is an example of a Probability Mass Function (PMF) because it calculates likelihoods for discrete random variables. It is an extension of the Bernoulli distribution that can model only 1 trial.

In this post, learn how to use the binomial distribution, its cumulative form, and when you can use it. I also include a binomial calculator that you can use with what you learn.

Note that this post focuses on how to use and graph the binomial distribution. If you want to learn how to calculate the probabilities by hand, please read Binomial Distribution Formula: Probability, Standard Deviation & Mean .

Binomial Probabilities

I’ll start by using statistical software to calculate the binomial probabilities and create distribution plots. This process will help you understand what you can learn from it.

Suppose you’re playing a game where rolling sixes on a die is really good. You want to know the probability of rolling exactly three sixes in ten die rolls. In this example, the number of events is 3 (X), the number of trials is 10 (N), and the probability (p) is 1/6 = 0.1667.

My software tells me that the likelihood is:

The binomial probability distribution calculates a likelihood of 0.155095 for rolling precisely three sixes in ten rolls.

That’s interesting but perhaps not so helpful by itself. We’re also interested in the chances for rolling other numbers of sixes. Seeing the distribution of probabilities for different numbers of sixes is much more helpful.

Binomial Distribution Graph

The binomial distribution graph is useful because it displays the probability of differing numbers of successes (Xs) out of the total number of trials (N). In the graph below, the distribution plot finds the likelihood of rolling exactly no sixes, 1 six, 2 sixes, 3 sixes, . . ., and up to 10 sixes in the ten die rolls. Using this approach, the binomial distribution graph covers the complete range of possible successes up to the total number of trials.

I like these graphs because they emphasize how we’re working with a distribution, and it’s easy to see which values happen more frequently.

In the chart, each bar represents the probability of rolling a specific number of sixes out of ten die rolls. The graph does not show the chances for seven and higher because the likelihoods of that many sixes in just ten rolls are too low to display on the chart.

The binomial distribution graph indicates the probability of rolling no sixes is about 16%. The highest chance is rolling one six (32%). Although, rolling two sixes occurs almost as frequently. Probabilities drop off quickly starting with three sixes. Additionally, the bar for three sixes matches our earlier result of 0.155095.

Related post : Understanding Probability Distributions

Binomial Cumulative Distribution Function

The binomial probability distribution is excellent for understanding the likelihood of obtaining an exact number of events (X) within a certain number of trials (N). However, many times you’re not interested in just one specific value for a binomial random variable. For example, in the die rolling example above, you might know from experience that rolling three or more sixes within ten rolls means you’re doing well. So, you actually want to learn the probability of rolling at least three sixes.

Let me introduce you to the binomial cumulative distribution function.

Technically, the binomial cumulative probability calculates the likelihood of obtaining less than or equal to X events in N trials. If you need to obtain a ≥ probability, use the inverse cumulative distribution. These days, statistical software will generally let you specify the direction of the cumulative function for the binomial distribution from the start. I’ll use the binomial distribution graph again to show you how it works.

For our example, we want to know the chances of rolling ≥ 3 sixes in 10 rolls. Below, the shaded region shows the inverse cumulative probability of rolling at least three sixes in ten die rolls.

The likelihood for rolling three or more sixes in ten rolls is 0.2249, not quite 1 in 4.

For a real-world example, see how I’ve used the binomial distribution to model the number of flu infections (X) for the vaccinated vs. unvaccinated over 20 years (N).

Learn more about Cumulative Distribution Functions: Uses, Graphs & vs PDF .

Binomial Distribution Assumptions and Notation

The binomial distribution models the probabilities for a binomial random variable having exactly X successes occurring in N trials. Your variable must satisfy the following requirements to be a binomial random variable. The binomial distribution is appropriate only for data that fulfill these assumptions.

• There must be only two possible outcomes per trial . For example, defective or not defective, sale or no sale, pass or fail, etc.
• The trials are independent . One trial’s outcome does not affect the subsequent trial. For instance, one coin toss doesn’t affect the result of the following coin toss. Learn more about Independent Events .
• The probability remains constant over time . In some areas, this assumption is true due to the physical characteristics of the process, such as coin tosses and die rolls. However, the probability won’t necessarily remain constant in other contexts. For example, the likelihood that a manufacturing process creates defective parts can change over time. If the probability can change, use the P chart ( a control chart ) to confirm this assumption.

Bernoulli Trials

Typically, you’ll use the binomial distribution when you have Bernoulli Trials, also known as Binomial Experiments. These trials involve binomial random variables that satisfactorily follow the assumptions above. In these trials, analysts label one of the possible outcomes as a success and the other outcome a failure.

A Bernoulli trial contains a set number of trials where the probability of a success is constant. The experiment counts the number of successes (X) out of the total number of trials (N).

You can think of the binomial probability distribution as modeling the number of successes (X) in a sample size of N.

Parameters and Notation

The binomial distribution has two parameters , n and p.

• n : the number of trials.
• p : the event or success probability.

You denote a binomial distribution as b(n,p).

Alternatively, you can write X∼b(n,p), which means that your binomial random variable X follows a binomial probability distribution with n trials and an event probability of p.

The previous examples assess probabilities corresponding with rolling sixes in a series of 10 die rolls. In this scenario, success is rolling a six, while a failure is rolling anything other than a six. The probability of rolling a six is 1/6 = 0.1667.

If rolling sixes is our random variable X, and we roll the die ten times, we can use the following notation for the binomial distribution:

X∼b(10,0.1667)

Binomial Distribution Calculator

Use this binomial distribution calculator to calculate the binomial probabilities and cumulative probabilities. Note that it uses “events” to indicate the number of trials (n).

Next, change exactly r successes to r or more successes . The calculator displays 22.487, matching the results for our example with the binomial inverse cumulative distribution.

Now, try one yourself. Imagine you’re drawing a random sample of 20 from a population where 10% are statisticians. You’re hoping that your study will have 3 or fewer statisticians because they’ll gang up and ask too many pesky questions about your study design. What is the likelihood of obtaining ≤ 3 statisticians?

See the correct answer at the end of this post.

Finally, the binomial and beta distributions are closely related. Click the link to learn more!

For more information about how to use binary data, read my posts, Maximize the Value of Your Binary Data , the Negative Binomial Distribution , the Geometric Distribution , and the Hypergeometric Distribution .

In the calculator example, there is an 86.7% chance of having ≤ 3 statisticians in your sample of 20 people.

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Reader Interactions

August 25, 2022 at 9:55 pm

Is there some way to combine binomial distributions? Here’s an example. Ann, Bob, and Carol are shooting threes on a basketball court. Ann takes 50 shots and has a 30% success rate. Bob takes 30 shots and has a 20% success rate. Carol takes 20 shots and has a 10% success rate. I can use the cumulative binomial distribution to calculate the chance that Ann makes 10 or more shots or that Bob makes 10 or more shots. How do I calculate the probability that the three of them combine to make 20 or more shots?

August 25, 2022 at 6:45 pm

Would binomial distributions be suitable for determining the probability of a prisoner re-offending once released from prison? Thank you.

Comments and Questions Cancel reply

4.3 Binomial Distribution

There are three characteristics of a binomial experiment.

• There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter n denotes the number of trials.
• There are only two possible outcomes, called "success" and "failure," for each trial. The letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on one trial. p + q = 1.
• The n trials are independent and are repeated using identical conditions. Because the n trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. Another way of saying this is that for each individual trial, the probability, p , of a success and probability, q , of a failure remain the same. For example, randomly guessing at a true-false statistics question has only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Joe always guesses correctly on any statistics true-false question with probability p = 0.6. Then, q = 0.4. This means that for every true-false statistics question Joe answers, his probability of success ( p = 0.6) and his probability of failure ( q = 0.4) remain the same.

The outcomes of a binomial experiment fit a binomial probability distribution . The random variable X = the number of successes obtained in the n independent trials.

The mean, μ , and variance, σ 2 , for the binomial probability distribution are μ = np and σ 2 = npq . The standard deviation, σ , is then σ = n p q n p q .

Any experiment that has characteristics two and three and where n = 1 is called a Bernoulli Trial (named after Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials.

Example 4.9

At ABC College, the withdrawal rate from an elementary physics course is 30% for any given term. This implies that, for any given term, 70% of the students stay in the class for the entire term. A "success" could be defined as an individual who withdrew. The random variable X = the number of students who withdraw from the randomly selected elementary physics class.

The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offer fruit in their lunches every day. This implies that 52% do not. What would a "success" be in this case?

Example 4.10

Suppose you play a game that you can only either win or lose. The probability that you win any game is 55%, and the probability that you lose is 45%. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times. Here, if you define X as the number of wins, then X takes on the values 0, 1, 2, 3, ..., 20. The probability of a success is p = 0.55. The probability of a failure is q = 0.45. The number of trials is n = 20. The probability question can be stated mathematically as P ( x = 15).

Try It 4.10

A trainer is teaching a rescued dolphin to catch live fish before returning it to the wild. The probability that the dolphin successfully catches a fish is 35%, and the probability that the dolphin does not successfully catch the fish is 65%. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. State the probability question mathematically.

Example 4.11

A coin is has been altered to weight the outcome from 0.5 to 0.25 and flipped 5 times. Each flip is independent. What is the probability of getting more than 3 heads? Let X = the number of heads in 5 flips of the fair coin. X takes on the values 0, 1, 2, 3, 4, 5. Since the coin is altered to result in p = 0.25, q is 0.75. The number of trials is n = 5. State the probability question mathematically.

First develop fully the probability density function and graph the probability density function. With the fully developed probability density function we can simply read the solution to the question P x > 3 P x > 3 heads. P x > 3 = P x = 4 + P x = 5 = 0 . 0146 + 0 . 0007 = 0 . 0153 . P x > 3 = P x = 4 + P x = 5 = 0 . 0146 + 0 . 0007 = 0 . 0153 . We have added the two individual probabilities because of the addition rule from Probability Topics .

Figure 4.2 also allows us to see the link between the probability density function and probability and area. We also see in Figure 4.2 the skew of the binomial distribution when p is not equal to 0.5. In Figure 4.2 the distribution is skewed right as a result of μ = n p = 1 . 25 μ = n p = 1 . 25 because p = 0 . 25 p = 0 . 25 .

Try It 4.11

A fair, six-sided die is rolled ten times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically.

Example 4.12

Approximately 70% of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly.

a. This is a binomial problem because there is only a success or a __________, there are a fixed number of trials, and the probability of a success is 0.70 for each trial.

b. If we are interested in the number of students who do their homework on time, then how do we define X ?

c. What values does x take on?

d. What is a "failure," in words?

e. If p + q = 1, then what is q ?

f. The words "at least" translate as what kind of inequality for the probability question P ( x ____ 40).

b. X = the number of statistics students who do their homework on time

c. 0, 1, 2, …, 50

d. Failure is defined as a student who does not complete their homework on time.

The probability of a success is p = 0.70. The number of trials is n = 50.

e. q = 0.30

f. greater than or equal to (≥) The probability question is P ( x ≥ 40).

Try It 4.12

Sixty-five percent of people pass the state driver’s exam on the first try. A group of 50 individuals who have taken the driver’s exam is randomly selected. Give two reasons why this is a binomial problem.

Notation for the Binomial: B = Binomial Probability Distribution Function

X ~ B ( n , p )

Read this as " X is a random variable with a binomial distribution." The parameters are n and p ; n = number of trials, p = probability of a success on each trial.

Example 4.13

It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education?

Let X = the number of workers who have a high school diploma but do not pursue any further education.

X takes on the values 0, 1, 2, ..., 20 where n = 20, p = 0.41, and q = 1 – 0.41 = 0.59. X ~ B (20, 0.41)

Find P ( x ≤ 12). P ( x ≤ 12) = 0.9738. (calculator or computer)

Using the TI-83, 83+, 84, 84+ Calculator

Go into 2 nd DISTR. The syntax for the instructions are as follows:

To calculate ( x = value): binompdf( n , p , number) if "number" is left out, the result is the binomial probability table. To calculate P ( x ≤ value): binomcdf( n , p , number) if "number" is left out, the result is the cumulative binomial probability table. For this problem: After you are in 2 nd DISTR , arrow down to binomcdf . Press ENTER . Enter 20,0.41,12). The result is P ( x ≤ 12) = 0.9738.

If you want to find P ( x = 12), use the pdf (binompdf). If you want to find P ( x > 12), use 1 - binomcdf(20,0.41,12).

The probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738.

The graph of X ~ B (20, 0.41) is as follows:

The y -axis contains the probability of x , where X = the number of workers who have only a high school diploma.

The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, μ = np = (20)(0.41) = 8.2.

The formula for the variance is σ 2 = npq . The standard deviation is σ = n p q n p q . σ = ( 20 ) ( 0.41 ) ( 0.59 ) ( 20 ) ( 0.41 ) ( 0.59 ) = 2.20.

Try It 4.13

About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator to find the answer.

Example 4.14

In the 2013 Jerry’s Artarama art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X = the number of pages that feature signature artists.

• What values does x take on?
• the probability that two pages feature signature artists
• the probability that at most six pages feature signature artists
• the probability that more than three pages feature signature artists.
• Using the formulas, calculate the (i) mean and (ii) standard deviation.
• x = 0, 1, 2, 3, 4, 5, 6, 7, 8
• P ( x = 2) = binompdf ( 100 , 8 560 , 2 ) ( 100 , 8 560 , 2 ) = 0.2466
• P ( x ≤ 6) = binomcdf ( 100 , 8 560 , 6 ) ( 100 , 8 560 , 6 ) = 0.9994
• P ( x > 3) = 1 – P ( x ≤ 3) = 1 – binomcdf ( 100 , 8 560 , 3 ) ( 100 , 8 560 , 3 ) = 1 – 0.9443 = 0.0557
• Mean = np = (100) ( 8 560 ) ( 8 560 ) = 800 560 800 560 ≈ 1.4286
• Standard Deviation = n p q n p q = ( 100 ) ( 8 560 ) ( 552 560 ) ( 100 ) ( 8 560 ) ( 552 560 ) ≈ 1.1867

Try It 4.14

According to a Gallup poll, 60% of American adults prefer saving over spending. Let X = the number of American adults out of a random sample of 50 who prefer saving to spending.

• What is the probability distribution for X ?
• the probability that 25 adults in the sample prefer saving over spending
• the probability that at most 20 adults prefer saving
• the probability that more than 30 adults prefer saving
• Using the formulas, calculate the (i) mean and (ii) standard deviation of X .

Example 4.15

The lifetime risk of developing cancer is about one in 67 (1.5%). Suppose we randomly sample 200 people. Let X = the number of people who will develop cancer.

• Use your calculator to find the probability that at most eight people develop cancer
• Is it more likely that five or six people will develop cancer? Justify your answer numerically.
• X   ~   B 200 , 0 . 015 X   ~   B 200 , 0 . 015
• Mean = n p = 200 0 . 015   = 3 Mean = n p = 200 0 . 015   = 3 Standard   Deviation = n p q = 200 ( 0 . 015 ) ( 0 . 985 ) = 1 . 719 Standard   Deviation = n p q = 200 ( 0 . 015 ) ( 0 . 985 ) = 1 . 719
• P x ≤ 8   = 0 . 9965 P x ≤ 8   = 0 . 9965
• The probability that five people develop cancer is 0.1011. The probability that six people develop cancer is 0.0500.

Try It 4.15

During a certain NBA season, a player for the Los Angeles Clippers had the highest field goal completion rate in the league. This player scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by this player during the season. Let X = the number of shots that scored points.

• Use your calculator to find the probability that this player scored with 60 of these shots.
• Find the probability that this player scored with more than 50 of these shots.

Example 4.16

The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC College has a student advisory committee made up of ten staff members and six students. The committee wishes to choose a chairperson and a recorder. What is the probability that the chairperson and recorder are both students? The names of all committee members are put into a box, and two names are drawn without replacement . The first name drawn determines the chairperson and the second name the recorder. There are two trials. However, the trials are not independent because the outcome of the first trial affects the outcome of the second trial. The probability of a student on the first draw is 6 16 6 16 . The probability of a student on the second draw is 5 15 5 15 , when the first draw selects a student. The probability is 6 15 6 15 , when the first draw selects a staff member. The probability of drawing a student's name changes for each of the trials and, therefore, violates the condition of independence.

Try It 4.16

A lacrosse team is selecting a captain. The names of all the seniors are put into a hat, and the first three that are drawn will be the captains. The names are not replaced once they are drawn (one person cannot be two captains). You want to see if the captains all play the same position. State whether this is binomial or not and state why.

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Statistics and probability

Course: statistics and probability   >   unit 9.

• Binomial variables
• Recognizing binomial variables
• 10% Rule of assuming "independence" between trials
• Identifying binomial variables

Binomial distribution

• Visualizing a binomial distribution
• Binomial probability example
• Generalizing k scores in n attempts
• Free throw binomial probability distribution
• Graphing basketball binomial distribution
• Binompdf and binomcdf functions
• Binomial probability (basic)
• Binomial probability formula
• Calculating binomial probability

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Video transcript

• Math Article

Binomial Distribution

In probability theory and statistics, the binomial distribution is the discrete probability distribution that gives only two possible results in an experiment, either Success or Failure . For example, if we toss a coin, there could be only two possible outcomes: heads or tails, and if any test is taken, then there could be only two results: pass or fail. This distribution is also called a binomial probability distribution.

There are two parameters n and p used here in a binomial distribution. The variable ‘n’ states the number of times the experiment runs and the variable ‘p’ tells the probability of any one outcome. Suppose a die is thrown randomly 10 times, then the probability of getting 2 for anyone throw is ⅙. When you throw the dice 10 times, you have a binomial distribution of n = 10 and p = ⅙. Learn the formula to calculate the two outcome distribution among multiple experiments along with solved examples here in this article.

Table of Contents:

Negative Binomial Distribution

• Mean and Variance

Binomial Distribution Vs Normal Distribution

• Solved Problems

Practice Problems

Binomial probability distribution.

In binomial probability distribution, the number of ‘Success’ in a sequence of n experiments, where each time a question is asked for yes-no, then the boolean-valued outcome is represented either with success/yes/true/one (probability p) or failure/no/false/zero (probability q = 1 − p). A single success/failure test is also called a Bernoulli trial or Bernoulli experiment, and a series of outcomes is called a Bernoulli process . For n = 1, i.e. a single experiment, the binomial distribution is a Bernoulli distribution . The binomial distribution is the base for the famous binomial test of statistical importance.

In probability theory and statistics, the number of successes in a series of independent and identically distributed Bernoulli trials before a particularised number of failures happens. It is termed as the negative binomial distribution. Here the number of failures is denoted by ‘r’. For instance, if we throw a dice and determine the occurrence of 1 as a failure and all non-1’s as successes. Now, if we throw a dice frequently until 1 appears the third time, i.e., r = three failures, then the probability distribution of the number of non-1s that arrived would be the negative binomial distribution.

Binomial Distribution Examples

As we already know, binomial distribution gives the possibility of a different set of outcomes. In real life, the concept is used for:

• Finding the quantity of raw and used materials while making a product.
• Taking a survey of positive and negative reviews from the public for any specific product or place.
• By using the YES/ NO survey, we can check whether the number of persons views the particular channel.
• To find the number of male and female employees in an organisation.
• The number of votes collected by a candidate in an election is counted based on 0 or 1 probability.

Also, read:

Binomial Distribution Formula

The binomial distribution formula is for any random variable X, given by;

n = the number of experiments

x = 0, 1, 2, 3, 4, …

p = Probability of Success in a single experiment

q = Probability of Failure in a single experiment = 1 – p

The binomial distribution formula can also be written in the form of n-Bernoulli trials, where n C x = n!/x!(n-x)!. Hence,

P(x:n,p) = n!/[x!(n-x)!].p x .(q) n-x

Binomial Distribution Mean and Variance

For a binomial distribution, the mean, variance and standard deviation for the given number of success are represented using the formulas

Mean, μ = np

Variance, σ 2 = npq

Standard Deviation σ= √(npq)

Where p is the probability of success

q is the probability of failure, where q = 1-p

The main difference between the binomial distribution and the normal distribution is that binomial distribution is discrete, whereas the normal distribution is continuous. It means that the binomial distribution has a finite amount of events, whereas the normal distribution has an infinite number of events. In case, if the sample size for the binomial distribution is very large, then the distribution curve for the binomial distribution is similar to the normal distribution curve.

Properties of Binomial Distribution

The properties of the binomial distribution are:

• There are two possible outcomes: true or false, success or failure, yes or no.
• There is ‘n’ number of independent trials or a fixed number of n times repeated trials.
• The probability of success or failure remains the same for each trial.
• Only the number of success is calculated out of n independent trials.
• Every trial is an independent trial, which means the outcome of one trial does not affect the outcome of another trial.

Binomial Distribution Examples And Solutions

Example 1: If a coin is tossed 5 times, find the probability of:

(a) Exactly 2 heads

(b) At least 4 heads.

(a) The repeated tossing of the coin is an example of a Bernoulli trial. According to the problem:

Number of trials: n=5

Probability of head: p= 1/2 and hence the probability of tail, q =1/2

For exactly two heads:

P(x=2) = 5 C2 p 2 q 5-2 = 5! / 2! 3! × (½) 2 × (½) 3

P(x=2) = 5/16

(b) For at least four heads,

x ≥ 4, P(x ≥ 4) = P(x = 4) + P(x=5)

P(x = 4) = 5 C4 p 4 q 5-4 = 5!/4! 1! × (½) 4 × (½) 1 = 5/32

P(x = 5) = 5 C5 p 5 q 5-5 = (½) 5 = 1/32

P(x ≥ 4) = 5/32 + 1/32 = 6/32 = 3/16

Example 2: For the same question given above, find the probability of:

a) Getting at most 2 heads

Solution: P (at most 2 heads) = P(X ≤ 2) = P (X = 0) + P (X = 1) + + P (X = 2)

P(X = 0) = (½) 5 = 1/32

P(X=1) = 5 C 1 (½) 5. = 5/32

P(x=2) = 5 C2 p 2 q 5-2 = 5! / 2! 3! × (½) 2 × (½) 3 = 5/16

P(X ≤ 2) = 1/32 + 5/32 + 5/16 = 1/2

A fair coin is tossed 10 times, what are the probability of getting exactly 6 heads and at least six heads.

Let x denote the number of heads in an experiment.

Here, the number of times the coin tossed is 10. Hence, n=10.

The probability of getting head, p ½

The probability of getting a tail, q = 1-p = 1-(½) = ½.

The binomial distribution is given by the formula:

P(X= x) = n C x p x q n-x , where = 0, 1, 2, 3, …

Therefore, P(X = x) = 10 C x (½) x (½) 10-x

(i) The probability of getting exactly 6 heads is:

P(X=6) = 10 C 6 (½) 6 (½) 10-6

P(X= 6) = 10 C 6 (½) 10

P(X = 6) = 105/512.

Hence, the probability of getting exactly 6 heads is 105/512.

(ii) The probability of getting at least 6 heads is P(X ≥ 6)

P(X ≥ 6) = P(X=6) + P(X=7) + P(X= 8) + P(X = 9) + P(X=10)

P(X ≥ 6) = 10 C 6 (½) 10 + 10 C 7 (½) 10  + 10 C 8 (½) 10  + 10 C 9 (½) 10  + 10 C 10 (½) 10

P(X ≥ 6) = 193/512.

Solve the following problems based on binomial distribution:

• The mean and variance of the binomial variate X are 8 and 4 respectively. Find P(X<3).
• The binomial variate X lies within the range {0, 1, 2, 3, 4, 5, 6}, provided that P(X=2) = 4P(x=4). Find the parameter “p” of the binomial variate X.
• In binomial distribution, X is a binomial variate with n= 100, p= ⅓, and P(x=r) is maximum. Find the value of r.

Frequently Asked Questions on Binomial Distribution

What is meant by binomial distribution.

The binomial distribution is the discrete probability distribution that gives only two possible results in an experiment, either success or failure.

Mention the formula for the binomial distribution.

The formula for binomial distribution is: P(x: n,p) = n C x p x (q) n-x Where p is the probability of success, q is the probability of failure, n= number of trials

What is the formula for the mean and variance of the binomial distribution?

The mean and variance of the binomial distribution are: Mean = np Variance = npq

What are the criteria for the binomial distribution?

The number of trials should be fixed. Each trial should be independent. The probability of success is exactly the same from one trial to the other trial.

What is the difference between a binomial distribution and normal distribution?

The binomial distribution is discrete, whereas the normal distribution is continuous.

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The Binomial Distribution

Tossing a Coin:

• Did we get Heads (H) or

We say the probability of the coin landing H is ½ And the probability of the coin landing T is ½

Throwing a Die:

• Did we get a four ... ?
• ... or not?

We say the probability of a four is 1/6 (one of the six faces is a four) And the probability of not four is 5/6 (five of the six faces are not a four)

Note that a die has 6 sides but here we look at only two cases: "four: yes" or "four: no"

Let's Toss a Coin!

Toss a fair coin three times ... what is the chance of getting exactly two Heads ?

Using H for heads and T for Tails we may get any of these 8 outcomes :

Which outcomes do we want?

"Two Heads" could be in any order: "HHT", "THH" and "HTH" all have two Heads (and one Tail).

So 3 of the outcomes produce "Two Heads".

What is the probability of each outcome?

Each outcome is equally likely, and there are 8 of them, so each outcome has a probability of 1/8

So the probability of event "Two Heads" is:

So the chance of getting Two Heads is 3/8

We used special words:

• Outcome : any result of three coin tosses (8 different possibilities)
• Event : "Two Heads" out of three coin tosses (3 outcomes have this)

3 Heads, 2 Heads, 1 Head, None

The calculations are (P means "Probability of"):

• P(Three Heads) = P( HHH ) = 1/8
• P(Two Heads) = P( HHT ) + P( HTH ) + P( THH ) = 1/8 + 1/8 + 1/8 = 3/8
• P(One Head) = P( HTT ) + P( THT ) + P( TTH ) = 1/8 + 1/8 + 1/8 = 3/8
• P(Zero Heads) = P( TTT ) = 1/8

We can write this in terms of a Random Variable "X" = "The number of Heads from 3 tosses of a coin":

• P(X = 3) = 1/8
• P(X = 2) = 3/8
• P(X = 1) = 3/8
• P(X = 0) = 1/8

And this is what it looks like as a graph:

Making a Formula

Now imagine we want the chances of 5 heads in 9 tosses : to list all 512 outcomes will take a long time!

So let's make a formula.

In our previous example, how can we get the values 1, 3, 3 and 1 ?

Well, they are actually in Pascal’s Triangle !

Can we make them using a formula?

Sure we can, and here it is:

The formula may look scary but is easy to use. We only need two numbers:

• n = total number
• k = number we want

The "!" means " factorial ", for example 4! = 1×2×3×4 = 24

Note: it is often called "n choose k" and you can learn more here .

Let's try it:

Example: with 3 tosses, what are the chances of 2 Heads?

We have n=3 and k=2 :

So there are 3 outcomes that have "2 Heads"

(We knew that already, but we now have a formula for it.)

Let's use it for a harder question:

Example: with 9 tosses, what are the chances of 5 Heads?

We have n=9 and k=5 :

So 126 of the outcomes will have 5 heads

And for 9 tosses there are a total of 2 9 = 512 outcomes, so we get the probability:

P(X=5)  =   126 512   = 0.24609375 

About a 25% chance .

(Easier than listing them all.)

So far the chances of success or failure have been equally likely .

But what if the coins are biased (land more on one side than another) or choices are not 50/50.

Example: You sell sandwiches. 70% of people choose chicken, the rest choose something else.

What is the probability of selling 2 chicken sandwiches to the next 3 customers.

This is just like the heads and tails example, but with 70/30 instead of 50/50.

Let's draw a tree diagram :

The "Two Chicken" cases are highlighted.

The probabilities for "two chickens" all work out to be 0.147 , because we are multiplying two 0.7s and one 0.3 in each case. In other words

0.147 = 0.7 × 0.7 × 0.3

Or, using exponents:

= 0.7 2 × 0.3 1

The 0.7 is the probability of each choice we want, call it p

The 2 is the number of choices we want, call it k

And we have (so far):

= p k × 0.3 1

The 0.3 is the probability of the opposite choice, so it is: 1−p

The 1 is the number of opposite choices, so it is: n−k

Which gives us:

= p k (1-p) (n-k)

• p is the probability of each choice we want
• k is the the number of choices we want
• n is the total number of choices

Example: (continued)

• p = 0.7 (chance of chicken)
• k = 2 (chicken choices)
• n = 3 (total choices)

which is what we got before, but now using a formula

Now we know the probability of each outcome is 0.147

But we need to include that there are three such ways it can happen: (chicken, chicken, other) or (chicken, other, chicken) or (other, chicken, chicken)

The total number of "two chicken" outcomes is:

And we get:

So the probability of event "2 people out of 3 choose chicken" = 0.441

OK. That was a lot of work for something we knew already, but now we have a formula we can use for harder questions.

Example: Sam says "70% choose chicken, so 7 of the next 10 customers should choose chicken" ... what are the chances Sam is right?

So we have:

That is the probability of each outcome.

And the total number of those outcomes is:

So the probability of 7 out of 10 choosing chicken is only about 27%

Moral of the story: even though the long-run average is 70%, don't expect 7 out of the next 10.

Putting it Together

Now we know how to calculate how many :

n! k!(n-k)!

And the probability of each :

p k (1-p) (n-k)

When multiplied together we get:

Probability of k out of n ways:

P(k out of n) =   n! k!(n-k)!   p k (1-p) (n-k)

The General Binomial Probability Formula

Important Notes:

• The trials are independent ,
• There are only two possible outcomes at each trial,
• The probability of "success" at each trial is constant.

Have a play with the Quincunx (then read Quincunx Explained ) to see the Binomial Distribution in action.

Throw the Die

A fair die is thrown four times. Calculate the probabilities of getting:

In this case n=4 , p = P(Two) = 1/6

X is the Random Variable ‘Number of Twos from four throws’.

Substitute x = 0 to 4 into the formula:

P(k out of n) =   n! k!(n-k)! p k (1-p) (n-k)

Like this (to 4 decimal places):

• P(X = 0) = 4! 0!4! × (1/6) 0 (5/6) 4 = 1 × 1 × (5/6) 4 = 0.4823
• P(X = 1) = 4! 1!3! × (1/6) 1 (5/6) 3 = 4 × (1/6) × (5/6) 3 = 0.3858
• P(X = 2) = 4! 2!2! × (1/6) 2 (5/6) 2 = 6 × (1/6) 2 × (5/6) 2 = 0.1157
• P(X = 3) = 4! 3!1! × (1/6) 3 (5/6) 1 = 4 × (1/6) 3 × (5/6) = 0.0154
• P(X = 4) = 4! 4!0! × (1/6) 4 (5/6) 0 = 1 × (1/6) 4 × 1 = 0.0008

Summary: "for the 4 throws, there is a 48% chance of no twos, 39% chance of 1 two, 12% chance of 2 twos, 1.5% chance of 3 twos, and a tiny 0.08% chance of all throws being a two (but it still could happen!)"

This time the graph is not symmetrical:

It is skewed because p is not 0.5

Sports Bikes

Your company makes sports bikes. 90% pass final inspection (and 10% fail and need to be fixed).

What is the expected Mean and Variance of the 4 next inspections?

First, let's calculate all probabilities.

• p = P(Pass) = 0.9

X is the Random Variable "Number of passes from four inspections".

• P(X = 0) = 4! 0!4! × 0.9 0 0.1 4 = 1 × 1 × 0.0001 = 0.0001
• P(X = 1) = 4! 1!3! × 0.9 1 0.1 3 = 4 × 0.9 × 0.001 = 0.0036
• P(X = 2) = 4! 2!2! × 0.9 2 0.1 2 = 6 × 0.81 × 0.01 = 0.0486
• P(X = 3) = 4! 3!1! × 0.9 3 0.1 1 = 4 × 0.729 × 0.1 = 0.2916
• P(X = 4) = 4! 4!0! × 0.9 4 0.1 0 = 1 × 0.6561 × 1 = 0.6561

Summary: "for the 4 next bikes, there is a tiny 0.01% chance of no passes, 0.36% chance of 1 pass, 5% chance of 2 passes, 29% chance of 3 passes, and a whopping 66% chance they all pass the inspection."

Mean, Variance and Standard Deviation

Let's calculate the Mean , Variance and Standard Deviation for the Sports Bike inspections.

There are (relatively) simple formulas for them. They are a little hard to prove, but they do work!

The mean, or "expected value", is:

For the sports bikes:

μ = 4 × 0.9 = 3.6

So we can expect 3.6 bikes (out of 4) to pass the inspection. Makes sense really ... 0.9 chance for each bike times 4 bikes equals 3.6

The formula for Variance is:

Variance: σ 2 = np(1-p)

And Standard Deviation is the square root of variance:

σ = √(np(1-p))

Variance: σ 2 = 4 × 0.9 × 0.1 = 0.36

Standard Deviation is:

σ = √(0.36) = 0.6

Note: we could also calculate them manually, by making a table like this:

The mean is the Sum of (X × P(X)) :

The variance is the Sum of (X 2 × P(X)) minus Mean 2 :

Variance: σ 2 = 13.32 − 3.6 2 = 0.36

• Mean value of X: μ = np
• Variance of X: σ 2 = np(1-p)
• Standard Deviation of X: σ = √(np(1-p))

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What Is Binomial Distribution?

• How It Works

The Bottom Line

• Corporate Finance
• Financial Analysis

Binomial Distribution: Definition, Formula, Analysis, and Example

Investopedia / Eliana Rodgers

Binomial distribution is a statistical distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters or assumptions.

The underlying assumptions of binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive or independent of one another.

Key Takeaways

• Binomial distribution is a statistical probability distribution that summarizes the likelihood that a value will take one of two independent values under a given set of parameters or assumptions.
• The underlying assumptions of binomial distribution are that there is only one outcome for each trial, that each trial has the same probability of success, and that each trial is mutually exclusive or independent of one another.
• Binomial distribution is a common discrete distribution used in statistics, as opposed to a continuous distribution, such as normal distribution.

Understanding Binomial Distribution

To start, the “binomial” in binomial distribution means two terms—the number of successes and the number of attempts. Each is useless without the other.

Binomial distribution is a common discrete distribution used in statistics, as opposed to a continuous distribution, such as normal distribution . This is because binomial distribution only counts two states, typically represented as 1 (for a success) or 0 (for a failure), given a number of trials in the data. Binomial distribution thus represents the probability for x successes in n trials, given a success probability p for each trial.

Binomial distribution summarizes the number of trials, or observations, when each trial has the same probability of attaining one particular value. Binomial distribution determines the probability of observing a specific number of successful outcomes in a specified number of trials.

Binomial distribution is often used in social science statistics as a building block for models for dichotomous outcome variables, such as whether a Republican or Democrat will win an upcoming election, whether an individual will die within a specified period of time, etc. It also has applications in finance, banking, and insurance, among other industries.

Analyzing Binomial Distribution

A binomial distribution's expected value, or mean, is calculated by multiplying the number of trials (n) by the probability of successes (p), or n × p.

For example, the expected value of the number of heads in 100 trials of heads or tails is 50, or (100 × 0.5). Another common example of binomial distribution is estimating the chances of success for a free-throw shooter in basketball, where 1 = a basket made and 0 = a miss.

The binomial distribution function is calculated as:

P ( x : n , p ) =  n C x p x ( 1 - p ) n - x

• n is the number of trials (occurrences)
• x is the number of successful trials
• p is the probability of success in a single trial
• n C x is the combination of n and x. A combination is the number of ways to choose a sample of x elements from a set of n distinct objects where order does not matter, and replacements are not allowed. Note that n C x = n! / r! ( n − r ) ! ), where ! is factorial (so, 4! = 4 × 3 × 2 × 1).

The mean of the binomial distribution is np, and the variance of the binomial distribution is np (1 − p). When p = 0.5, the distribution is symmetric around the mean—such as when flipping a coin because the chances of getting heads or tails is 50%, or 0.5. When p > 0.5, the distribution curve is skewed to the left. When p < 0.5, the distribution curve is skewed to the right.

The binomial distribution is the sum of a series of multiple independent and identically distributed Bernoulli trials. In a Bernoulli trial, the experiment is said to be random and can only have two possible outcomes: success or failure.

For instance, flipping a coin is considered to be a Bernoulli trial ; each trial can only take one of two values (heads or tails), each success has the same probability, and the results of one trial do not influence the results of another. Bernoulli distribution is a special case of binomial distribution where the number of trials n = 1.

Example of Binomial Distribution

Binomial distribution is calculated by multiplying the probability of success raised to the power of the number of successes and the probability of failure raised to the power of the difference between the number of successes and the number of trials. Then, multiply the product by the combination of the number of trials and successes.

For example, assume that a casino created a new game in which participants can place bets on the number of heads or tails in a specified number of coin flips. Assume a participant wants to place a $10 bet that there will be exactly six heads in 20 coin flips. The participant wants to calculate the probability of this occurring, and therefore, they use the calculation for binomial distribution.

The probability was calculated as (20! / (6! × (20 - 6)!)) × (0.50) (6) × (1 - 0.50) (20 - 6) . Consequently, the probability of exactly six heads occurring in 20 coin flips is 0.0369, or 3.7%. The expected value was 10 heads in this case, so the participant made a poor bet. The graph below shows that the mean is 10 (the expected value), and the chances of getting six heads is on the left tail in red. You can see that there is less of a chance of six heads occurring than seven, eight, nine, 10, 11, 12, or 13 heads.

StatCrunch Binomial Calculator

So how can this be used in finance? One example: Let’s say you’re a bank, a lender , that wants to know within three decimal places the likelihood of a particular borrower defaulting. What are the chances of so many borrowers defaulting that they would render the bank insolvent? Once you use the binomial distribution function to calculate that number, you have a better idea of how to price insurance and, ultimately, how much money to lend out and keep in reserve. 

Binomial distribution is a statistical probability distribution that states the likelihood that a value will take one of two independent values under a given set of parameters or assumptions.

How Is Binomial Distribution Used?

This distribution pattern is used in statistics but has implications in finance and other fields. Banks may use it to estimate the likelihood of a particular borrower defaulting, how much money to lend, and the amount to keep in reserve. It’s also used in the insurance industry to determine policy pricing and assess risk .

Why Is Binomial Distribution Important?

Binomial distribution is used to figure the probability of a pass or fail outcome in a survey, or experiment replicated numerous times. There are only two potential outcomes for this type of distribution. More broadly, distribution is an important part of analyzing data sets to estimate all the potential outcomes of the data and how frequently they occur. Forecasting and understanding the success or failure of outcomes is essential to business development.

The binomial distribution is an important statistical distribution that describes binary outcomes (such as the flip of a coin, a yes/no answer, or an on/off condition). Understanding its characteristics and functions is important for data analysis in various contexts that involve an outcome taking one of two independent values.

It has applications in social science, finance, banking, insurance, and other areas. For instance, it can be used to estimate whether a borrower will default on a loan, whether an options contract will finish in-the-money or out-of-the-money, or whether a company will miss or beat earnings estimates.

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Binomial Distribution

In statistics and probability theory, the binomial distribution is the probability distribution that is discrete and applicable to events having only two possible results in an experiment, either success or failure. (the prefix “bi” means two, or twice). A few circumstances where we have binomial experiments are tossing a coin: head or tail, the result of a test: pass or fail, selected in an interview: yes/ no, or nature of the product: defective/non-defective. Such a distribution of a binomial random variable is called a binomial probability distribution.

Binomial Distribution is a commonly used discrete distribution in statistics. The normal distribution as opposed to a binomial distribution is a continuous distribution. Let us learn the formula to calculate the Binomial distribution considering many experiments and a few solved examples for a better understanding.

What Is Binomial Distribution?

The binomial distribution is the probability distribution of a binomial random variable. A random variable is a real-valued function whose domain is the sample space of a random experiment. Let us consider an example to understand this better.

Toss a fair coin twice. This is a binomial experiment. There are 4 possible outcomes of this experiment. {HH, HT, TH, TT}. Consider getting one head as the success. Count the number of successes in each possible outcome. Here n(getting heads) is the success in n repeated trials of a binomial experiment.. n(X) = 0, 1, or 2 is the binomial random variable . The distribution of probability is of a binomial random variable, and this is known as a binomial distribution.

This table shows that getting one head in a single flip is 0.50. Now if a coin is flipped 3 times, consider we are intended to find the binomial distribution of getting two heads. Tossing 3 coins result in 8 outcomes. {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. The probability of getting two heads [P(HH)] is 3/8. Similarly, we can calculate the probability of getting one head, 2 heads, and 3 heads and 0 heads. The binomial probability distribution is given in terms of a random variable as:

P(X = 0) = 1/8

P(X = 1) = 3/8

P(X = 2) = 3/8

P(X = 3)= 1/8

Binomial Distribution in Statistics

The binomial distribution forms the base for the famous binomial test of statistical importance. The binomial distribution represents the probability for 'x' successes of an experiment in 'n' trials, given a success probability 'p' for each trial at the experiment. Two parameters n and p are used here in the binomial distribution. The variable ‘n’ represents the number of trials and the variable ‘p’ states the probability of any one(each) outcome. A test that has a single outcome such as success/failure is also called a Bernoulli trial or Bernoulli experiment, and a series of outcomes is called a Bernoulli process.

Consider an experiment where each time a question is asked for a yes/no with a series of n experiments. Then in the binomial probability distribution, the boolean-valued outcome the success/yes/true/one is represented with probability p and the failure/no/false/zero with probability q (q = 1 − p). In a single experiment when n = 1, the binomial distribution is called a Bernoulli distribution .

If a die is thrown randomly 10 times, then the probability of getting a 3 for any throw is 1/6. Similarly, if we throw the dice 10 times, we have n = 10 and p = 1/6, q = 5/6

Negative Binomial Distribution

Let's understand with an example when can a binomial distribution be negative. Suppose we throw a die and determine that the occurrence of 2 will be a failure and all non-2’s will be successes. Let the failures be denoted by ‘r’. Now, if the die is thrown frequently until 2 appears the third time, i.e., r = three failures, then the binomial distribution of the number of non-2's that arrived would be the negative binomial distribution .

Binomial Distribution Formula

The binomial distribution formula is for any random variable X, given by; P(x:n,p) = n C x p x (1-p) n-x Or P(x:n,p) = n C x p x (q) n-x

Where p is the probability of success, q is the probability of failure, and n = number of trials. The binomial distribution formula is also written in the form of n-Bernoulli trials.

where n C x = n!/x!(n-x)!. Hence, P(x:n,p) = n!/[x!(n-x)!].p x .(q) n-x

Binomial Distribution Calculation

The image given below shows the formula used for binomial distribution calculation:

Application of Binomial Distribution

We now already know that binomial distribution gives the probability of a different set of outcomes. In real life, the concept of the binomial distribution is used for:

• Finding the quantity of raw and used materials while making a product.
• Taking a survey of positive and negative reviews from the public for any specific product or place.
• By using the YES/ NO survey
• To find the number of male and female students in a university.
• The number of votes collected by a candidate in an election is counted based on 0 or 1 probability.

Consider a card selected at a random and replaced. If this experiment is repeated 5 times, let us find the probability of selecting exactly 3 hearts. Let us determine the number of trials, success, and the failure. The trial is the drawing a card 5 times. Thus n = 5.

success: card drawn is a heart = p = 1/4 = 0.25

failure: card drawn is not a heart = q = 1-0.25 = 0.75

Using the binomial distribution formula, we get 5 C \(_3\) (0,25) 3 (0.75) 2 = 0.088

Binomial Distribution Mean and Variance

For a binomial distribution, the mean , variance and standard deviation for the given number of success are represented using the formulas

• Mean, μ = np
• Variance, σ 2 = npq
• Standard Deviation σ= √(npq)

Where p is the probability of success q is the probability of failure, where q = 1-p

Binomial Distribution Vs Normal Distribution

The main difference between the binomial distribution and the normal distribution is that the binomial distribution is discrete, whereas the normal distribution is continuous. It means that the binomial distribution has a finite amount of events, whereas the normal distribution has an infinite number of events. In case, if the sample size for the binomial distribution is very large, then the distribution curve for the binomial distribution is similar to the normal distribution curve.

Properties of Binomial Distribution

The properties of the binomial distribution are:

• There are only two distinct possible outcomes: true/false, success/failure, yes/no.
• There is a fixed number of 'n' times repeated trials in a given experiment.
• The probability of success or failure remains constant for each attempt/trial.
• Only the successful attempts are calculated out of 'n' independent trials.
• Every trial is an independent trial on its own, this means that the outcome of one trial has no effect on the outcome of another trial.

Important Notes on Binomial Distribution

• For using the binomial distribution, the number of observations or trials in an experiment is fixed or finite.
• Each observation/attempt/trial is independent on its own. This means none of the trials have an effect on the probability of the next trial.
• Each trial has an equal probability of occurrence. The probability of success is exactly the same from one trial to another.

☛ Related Articles:

• Normal Distribution Formula
• Cumulative Frequency
• Frequency Distribution

Binomial Distribution Examples

Example 1: If a coin is tossed 5 times, using binomial distribution find the probability of:

(a)Exactly 2 heads

(b) At least 4 heads.

(a) The repeated tossing of the coin is an example of a Bernoulli trial. According to the problem:

Number of trials: n=5

Probability of head: p= 1/2 and hence the probability of tail, q =1/2

For exactly two heads:

Using binomial distribution formula,

P(x=2) = 5 C 2 p 2 q 5-2 = 5! / 2! 3! × (½) 2 × (½) 3

P(x=2) = 5/16

(b) For at least four heads,

x ≥ 4, P(x ≥ 4) = P(x = 4) + P(x=5)

Hence, using binomial distribution formula,

P(x = 4) = 5 C 4 p 4 q 5-4 = 5!/4! 1! × (½) 4 × (½) 1 = 5/32

P(x = 5) = 5 C 5 p 5 q 5-5 = (½) 5 = 1/32

Answer: Therefore, P(x ≥ 4) = 5/32 + 1/32 = 6/32 = 3/16

Example 2: For the same question given above, using the binomial distribution find the probability of getting at most 2 heads.

Solution: P(at most 2 heads) = P(X ≤ 2) = P (X = 0) + P (X = 1)

P(X = 0) = (½)5 = 1/32

Using binomial distribution formula, we get

P(X=1) = 5 C 1 (½) 5 = 5/32

Answer: Therefore, P(X ≤ 2) = 1/32 + 5/32 = 3/16

Example 3: A random variable X has the following binomial distribution. Determine P(X>6) and P(0<X<3)

This is a binomial distribution.

To find k. The sum of all the probabilities = 1

0 + k + 2k +2k + 3k + k 2 + 2k 2 + 7k 2 + k = 1

10k 2 + 8 k = 1

Solving for k , we get k = 0.1 and -1, We consider k = 0.1 as k = -1 makes the probability negative which is not possible.

i) P(X>6)= 7k 2 + k

7(0.1) 2 + 0.1

ii) P(0<X<3)

= k + 2k = 3k

Answer: P(X>6)= 0.17 and P(0<X<3) = 0.3

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Binomial Distribution Questions

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FAQs on Binomial Distribution

What is a binomial distribution.

The binomial distribution is a common discrete distribution used in statistics, as opposed to a continuous distribution, such as the normal distribution . The binomial distribution, therefore, represents the probability for x successes in n trials, given a success probability p for each trial, and is applicable to events having only two possible results in an experiment.

What Is the Purpose of Binomial Distribution?

The binomial distribution model allows us to compute the probability of observing a specified number of "successes" when the process is repeated a specific number of times (e.g., in a set of patients) and the outcome for a given patient is either a success or a failure.

What Is the Formula for Binomial Distribution?

The formula for binomial distribution is:

P(x: n,p) = n C x p x (q) n-x

• n = the number of experiments
• x = 0, 1, 2, 3, 4, …
• p = Probability of success in a single experiment
• q = Probability of failure in a single experiment (= 1 – p)

What Is the Formula for the Mean and Variance of the Binomial Distribution?

The mean and variance of the binomial distribution are:

• Variance = npq

Where p is the probability of success, q is the probability of failure, and n = number of trials.

What Are the Criteria for the Binomial Distribution?

The criteria for using the binomial distribution are:

• The number of trials should be fixed.
• Each trial should be independent.
• The probability of success is exactly the same from one trial to the other trial.

What Is the Difference Between a Binomial Distribution and Normal Distribution?

How do you identify a binomial distribution.

For a variable to be a binomial random variable, all of the following conditions must be met:

• There are a fixed number of trials (a fixed sample size).
• On each trial, the event of interest either occurs or does not.
• The probability of occurrence (or not) is the same on each trial.
• Trials are independent of one another.

Is Binomial Distribution Discrete or Continuous?

A binomial distribution is a discrete distribution with parameters n and p, where n is the number of trials and p is the probability of success.

For the Binomial Distribution, Which Formula Finds the Standard Deviation?

The standard deviation formula for a binomial distribution is given by, σ = √(npq), where n = number of trials, p = probability of success, q = probability of failure = 1 - p.

What is Negative Binomial Distribution?

Negative binomial distribution is a discrete probability distribution in statistics. It helps in finding r success in x trials. Here we consider the n + r trials needed to get r successes.

Binomial Distribution Calculator

What is the binomial probability, binomial probability formula, how to use the binomial distribution calculator: an example, how to calculate cumulative probabilities, binomial probability distribution experiments, mean and variance of binomial distribution, other considerations.

This binomial distribution calculator is here to help you with probability problems in the following form: what is the probability of a certain number of successes in a sequence of events? Read on to learn what exactly is the binomial probability distribution, when and how to apply it, and learn the binomial probability formula. Find out what is binomial distribution, and discover how binomial experiments are used in various settings.

Imagine you're playing a game of dice. To win, you need exactly three out of five dice to show a result equal to or lower than 4. The remaining two dice need to show a higher number. What is the probability of you winning?

This is a sample problem that can be solved with our binomial probability calculator. You know the number of events (it is equal to the total number of dice, so five); you know the number of successes you need (precisely 3); you also can calculate the probability of one single success occurring (4 out of 6, so 0.667). This is all the data required to find the binomial probability of you winning the game of dice.

Note that to use the binomial distribution calculator effectively, the events you analyze must be independent . It means that all the trials in your example are supposed to be mutually exclusive.

The first trial's success doesn't affect the probability of success or the probability of failure in subsequent events, and they stay precisely the same. In the case of a dice game, these conditions are met: each time you roll a die constitutes an independent event.

Sometimes you may be interested in the number of trials you need to achieve a particular outcome. For instance, you may wonder how many rolls of a die are necessary before you throw a six three times. Such questions may be addressed using a related statistical tool called the negative binomial distribution. Make sure to learn about it with Omni's negative binomial distribution calculator .

Also, you may check our normal approximation to binomial distribution calculator and the related continuity correction calculator.

To find this probability, you need to use the following equation:

P(X=r) = nCr × p r × (1-p) n-r

• n – Total number of events;
• r – Number of required successes;
• p  – Probability of one success;
• nCr – Number of combinations (so-called "n choose r"); and
• P(X=r) – Probability of an exact number of successes happening.

You should note that the result is the probability of an exact number of successes. For example, in our game of dice, we needed precisely three successes – no less, no more. What would happen if we changed the rules so that you need at least three successes? Well, you would have to calculate the probability of exactly three, precisely four, and precisely five successes and sum all of these values together.

Let's solve the problem of the game of dice together.

Determine the number of events. n is equal to 5, as we roll five dice.

Determine the required number of successes. r is equal to 3, as we need exactly three successes to win the game.

The probability of rolling 1, 2, 3, or 4 on a six-sided die is 4 out of 6, or 0.667. Therefore p is equal to 0.667 or 66.7%.

Calculate the number of combinations (5 choose 3). You can use the combination calculator to do it. This number, in our case, is equal to 10.

Substitute all these values into the binomial probability formula above:

P(X = 3) = 10 × 0.667 3 × (1-0.667) (5-3) = 10 × 0.667 3 × (1-0.667) (5-3) = 10 × 0.296 × 0.333 2 = 2.96 × 0.111 = 0.329

You can also save yourself some time and use the binomial distribution calculator instead :)

Sometimes, instead of an exact number of successes, you want to know the probability of getting r or more successes or r or less successes. To calculate the probability of getting any range of successes:

• Use the binomial probability formula to calculate the probability of success (P) for all possible values of r you are interested in.
• Sum the values of P for all r within the range of interest.

For example, the probability of getting two or fewer successes when flipping a coin four times (p = 0.5 and n = 4) would be:

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X ≤ 2) = 37.5% + 25% + 6.25%

P(X ≤ 2) = 68.75%

This calculation is made easy using the options available on the binomial distribution calculator. You can change the settings to calculate the probability of getting:

• Exactly r successes: P(X = r)
• r or more successes: P(X ≥ r)
• r or fewer successes: P(X ≤ r)
• Between r₀ and r₁ successes P(r₀ ≤ X ≤ r₁)

The binomial distribution turns out to be very practical in experimental settings . However, the output of such a random experiment needs to be binary : pass or failure, present or absent, compliance or refusal. It's impossible to use this design when there are three possible outcomes.

At the same time, apart from rolling dice or tossing a coin, it may be employed in somehow less clear cases. Here are a couple of questions you can answer with the binomial probability distribution:

• Will a new drug work on a randomly selected patient?
• Will a light bulb you just bought work properly, or will it be broken?
• What is a chance of correctly answering a test question you just drew?
• What is a probability of a random voter to vote for a candidate in an election?
• How likely is it for a group of students to be accepted to a prestigious college?

Experiments with precisely two possible outcomes, such as the ones above, are typical binomial distribution examples, often called the Bernoulli trials .

In practice, you can often find the binomial probability examples in fields like quality control , where this method is used to test the efficiency of production processes. The inspection process based on the binomial distribution is designed to perform a sufficient number of checkups and minimize the chances of manufacturing a defective product.

If you don't know the probability of an independent event in your experiment ( p ), collect the past data in one of your binomial distribution examples, and divide the number of successes ( y ) by the overall number of events p = y/n .

Once you have determined your rate of success (or failure) in a single event, you need to decide what's your acceptable number of successes (or failures) in the long run. For example, one defective product in a batch of fifty is not a tragedy, but you wouldn't like to have every second product faulty, would you?

Bernoulli trials are also perfect at solving network systems . Interestingly, they may be used to work out paths between two nodes on a diagram. This is the case of the Wheatstone bridge network, a representation of a circuit built for electrical resistance measurement.

Like the binomial distribution table , our calculator produces results that help you assess the chances that you will meet your target. However, if you like, you may take a look at this binomial distribution table . It tells you what is the binomial distribution value for a given probability and number of successes.

One of the most exciting features of binomial distributions is that they represent the sum of a number n of independent events. Each of them ( Z ) may assume the values of 0 or 1 over a given period.

Let's say the probability that each Z occurs is p . Since the events are not correlated, we can use random variables' addition properties to calculate the mean (expected value) of the binomial distribution μ = np .

The variance of a binomial distribution is given as: σ² = np(1-p) . The larger the variance, the greater the fluctuation of a random variable from its mean. A small variance indicates that the results we get are spread out over a narrower range of values.

The standard deviation of binomial distribution, another measure of a probability distribution dispersion, is simply the square root of the variance, σ . Keep in mind that the standard deviation calculated from your sample (the observations you actually gather) may differ from the entire population's standard deviation. If you find this distinction confusing, there here's a great explanation of this distinction .

There's a clear-cut intuition behind these formulas. Suppose this time that I flip a coin 20 times:

• My p is then equal to 0.5 (unless, of course, the coin is rigged);
• Each Z has an equivalent chance of 0 or 1;
• The number of trials, n , is 20.

This sequence of events fulfills the prerequisites of a binomial distribution.

The mean value of this simple experiment is: np = 20 × 0.5 = 10 . We can say that on average if we repeat the experiment many times, we should expect heads to appear ten times.

The variance of this binomial distribution is equal to np(1-p) = 20 × 0.5 × (1-0.5) = 5 . Take the square root of the variance, and you get the standard deviation of the binomial distribution, 2.24 . Accordingly, the typical results of such an experiment will deviate from its mean value by around 2. Hence, in most of the trials, we expect to get anywhere from 8 to 12 successes.

Use our binomial probability calculator to get the mean, variance, and standard deviation of binomial distribution based on the number of events you provided and the probability of one success.

Developed by a Swiss mathematician Jacob Bernoulli , the binomial distribution is a more general formulation of the Poisson distribution. In the latter, we simply assume that the number of events (trials) is enormous, but the probability of a single success is small.

The binomial distribution is closely related to the binomial theorem , which proves to be useful for computing permutations and combinations. Make sure to check out our permutations calculator , too!

Keep in mind that the binomial distribution formula describes a discrete distribution . The possible outcomes of all the trials must be distinct and non-overlapping. What's more, the two outcomes of an event must be complementary: for a given p , there's always an event of q = 1-p .

If there's a chance of getting a result between the two, such as 0.5, the binomial distribution formula should not be used. The same goes for the outcomes that are non-binary, e.g., an effect in your experiment may be classified as low, moderate, or high.

However, for a sufficiently large number of trials, the binomial distribution formula may be approximated by the Gaussian (normal) distribution specification, with a given mean and variance. That allows us to perform the so-called continuity correction , and account for non-integer arguments in the probability function.

Maybe you still need some practice with the binomial probability distribution examples?

Try to solve the dice game's problem again, but this time you need three or more successes to win it. How about the chances of getting exactly 4?

Is the binomial distribution discrete or continuous?

The binomial distribution is discrete – it takes only a finite number of values.

How do I find the mean of a binomial distribution?

To calculate the mean (expected value) of a binomial distribution B(n,p) you need to multiply the number of trials n by the probability of successes p , that is: mean = n × p .

How do I find the standard deviation of a binomial distribution?

To find the standard deviation of a binomial distribution B(n,p) :

• Compute the variance as n × p × (1-p) , where n is the number of trials and p is the probability of successes.
• Take the square root of the number obtained in Step 1.
• That's it! Congrats :)

What is the probability of 3 successes in 5 trials if the probability of success is 0.5?

To find this probability, you need to:

Recall the binomial distribution formula P(X = r) = nCr × pʳ × (1-p)ⁿ⁻ʳ . We'll use it with the following data:

Number of trials: n = 5 ;

Number of successes: r = 3 ; and

Probability of success: p = 0.5 .

Calculate 5 choose 3 : nCr = 10 .

Plug these values into the formula:

P(X = 3) = 10 × 0.5² × 0.5³ = 0.3125 .

The probability you're looking for is 31.25% .

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Binomial Experiment: Rules, Examples, Steps

What is a binomial experiment.

A binomial experiment is an experiment where you have a fixed number of independent trials with only have two outcomes. For example, the outcome might involve a yes or no answer. If you toss a coin you might ask yourself “Will I get a heads?” and the answer is either yes or no. That’s the basic idea, but in order to call an experiment a binomial experiment you also have to make sure of the following rules.

• You must have a fixed number of trials . This should go without saying; if you don’t have a fixed number of trials you could be tossing that coin forever without stopping. In addition, the results from your experiment will be vastly different if you toss that coin twice (you could get two heads in a row and conclude that you will always get a heads if you toss a coin!) or if you toss it a hundred times .
• Each trial is an independent event . “Independent” means that every time you repeat the trial (i.e. tossing that coin), it’s a fresh new trial and nothing you do has an effect on each coin toss. For example, if you tossed ten coins at a time and removed the coins that landed heads down before throwing again, you’ll affect the probability, because there are fewer coins. There’s nothing wrong with that, but it would not be a binomial experiment. The fact that each trial is independent of each other leads to another important aspect of binomial experiments; the probability remains constant from trial to trial.
• There are only two outcomes. In other words, if you can phrase the experiment as a yes or no answer, then it can be a binomial experiment: Will I get a heads? Can someone find a parking space in the city? Do eggs hard boil in ten minutes?

Binomial Experiment: Examples

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• Tossing a coin a hundred times to see how many land on heads.
• Asking 100 people if they have ever been to Paris.
• Rolling two dice to see if you get a double.

Examples of experiments that are not Binomial Experiments

• Asking 100 people how much they weigh (you’ll have a hundred possible answers, not two).
• Tossing a coin until you get a heads (it could take one toss, or three, or six, so there is not a fixed number of trials). This is actually called a negative binomial experiment .

Binomial experiment: Four Steps

Determining if a question concerns a binomial experiment involves asking yourself four questions about the problem.

Example question: which of the following are binomial experiments?

• Telephone surveying a group of 200 people to ask if they voted for George Bush.
• Counting the average number of dogs seen at a veterinarian’s office daily.
• You take a survey of 50 traffic lights in a certain city, at 3 p.m., recording whether the light was red, green, or yellow at that time.
• You are at a fair, playing “pop the balloon” with 6 darts. There are 20 balloons. 10 of the balloons have a ticket inside that say “win,” and 10 have a ticket that says “lose.”

Step 1: Ask yourself: is there a fixed number of trials ?

• For question #1, the answer is yes (200).
• For question #2, the answer is no , so we’re going to discard #2 as a binomial experiment.
• For question #3, the answer is yes , there’s a fixed number of trials (the 50 traffic lights).
• For question #4, the answer is yes (your 6 darts).

Step 2: Ask yourself: Are there only 2 possible outcomes?

• For question #1, the only two possible outcomes are that they did, or they didn’t vote for Mr. Bush, so the answer is yes .
• For question #3, there are 3 possibilities: red, green, and yellow, so it’s not a binomial experiment.
• For question #4, the only possible outcomes are WIN or LOSE, so the answer is yes .

Step 3:  Ask yourself: are the outcomes independent of each other ? In other words, does the outcome of one trial (or one toss, or one question) affect another trial?

• For question #1, the answer is yes : one person saying they did or didn’t vote for Mr. Bush isn’t going to affect the next person’s response.
• For question #4, each time you toss a dart, the number of winning and losing tickets changes, which means, for example, if you win one toss, the probability of winning isn’t 10 to 10 anymore, but 9 to 10, since you already have one of the winning tickets. Since the probability is different, the trials are not independent events , so the answer is no , and question #4 is not a binomial experiment.

Step 4: Does the probability of success remain the same for each trial ?

• For question #1, the answer is yes , each person has a 50% chance of having voted for Mr. Bush.

Question #1 out of the 4 given questions was the only one that was a binomial experiment .

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Gonick, L. (1993). The Cartoon Guide to Statistics . HarperPerennial. Kotz, S.; et al., eds. (2006), Encyclopedia of Statistical Sciences , Wiley.

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Binomial Probability Calculator

Use the Binomial Calculator to compute individual and cumulative binomial probabilities. For help in using the calculator, read the Frequently-Asked Questions or review the Sample Problems .

To learn more about the binomial distribution, go to Stat Trek's tutorial on the binomial distribution .

• Enter a value in each of the first three text boxes (the unshaded boxes).
• Click the Calculate button to compute binomial and cumulative probabilities.

Frequently-Asked Questions

Instructions: To find the answer to a frequently-asked question, simply click on the question.

What is a binomial experiment?

A binomial experiment has the following characteristics:

• The experiment involves repeated trials.
• Each trial has only two possible outcomes - a success or a failure.
• The probability that any trial will result in success is constant.
• All of the trials in the experiment are independent.

A series of coin tosses is a perfect example of a binomial experiment. Suppose we toss a coin three times. Each coin flip represents a trial, so this experiment would have 3 trials. Each coin flip also has only two possible outcomes - a Head or a Tail. We could call a Head a success; and a Tail, a failure. The probability of a success on any given coin flip would be constant (i.e., 50%). And finally, the outcome on any coin flip is not affected by previous or succeeding coin flips; so the trials in the experiment are independent.

What is a binomial distribution?

A binomial distribution is a probability distribution . It refers to the probabilities associated with the number of successes in a binomial experiment .

For example, suppose we toss a coin three times and suppose we define Heads as a success. This binomial experiment has four possible outcomes: 0 Heads, 1 Head, 2 Heads, or 3 Heads. The probabilities associated with each possible outcome are an example of a binomial distribution , as shown below.

What is the number of trials?

The number of trials refers to the number of replications in a binomial experiment.

Suppose that we conduct the following binomial experiment. We flip a coin and count the number of Heads. We classify Heads as success; tails, as failure. If we flip the coin 3 times, then 3 is the number of trials. If we flip it 20 times, then 20 is the number of trials.

Note: Each trial results in a success or a failure. So the number of trials in a binomial experiment is equal to the number of successes plus the number of failures.

What is the number of successes?

Each trial in a binomial experiment can have one of two outcomes. The experimenter classifies one outcome as a success; and the other, as a failure. The number of successes in a binomial experient is the number of trials that result in an outcome classified as a success.

What is the probability of success on a single trial?

In a binomial experiment, the probability of success on any individual trial is constant. For example, the probability of getting Heads on a single coin flip is always 0.50. If "getting Heads" is defined as success, the probability of success on a single trial would be 0.50.

What is the binomial probability?

A binomial probability refers to the probability of getting EXACTLY r successes in a specific number of trials. For instance, we might ask: What is the probability of getting EXACTLY 2 Heads in 3 coin tosses. That probability (0.375) would be an example of a binomial probability.

In a binomial experiment, the probability that the experiment results in exactly x successes is indicated by the following notation: P(X=x);

What is the cumulative binomial probability?

Cumulative binomial probability refers to the probability that the value of a binomial random variable falls within a specified range.

The probability of getting AT MOST 2 Heads in 3 coin tosses is an example of a cumulative probability. It is equal to the probability of getting 0 heads (0.125) plus the probability of getting 1 head (0.375) plus the probability of getting 2 heads (0.375). Thus, the cumulative probability of getting AT MOST 2 Heads in 3 coin tosses is equal to 0.875.

Notation associated with cumulative binomial probability is best explained through illustration. The probability of getting FEWER THAN 2 successes is indicated by P(X<2); the probability of getting AT MOST 2 successes is indicated by P(X≤2); the probability of getting AT LEAST 2 successes is indicated by P(X≥2); the probability of getting MORE THAN 2 successes is indicated by P(X>2).

Sample Problem

• The probability of success (i.e., getting a Head) on any single trial is 0.5.
• The number of trials is 12.
• The number of successes is 7 (since we define getting a Head as success).

Therefore, we plug those numbers into the Binomial Calculator and hit the Calculate button.

The calculator reports that the binomial probability is 0.193. That is the probability of getting EXACTLY 7 Heads in 12 coin tosses. (The calculator also reports the cumulative probabilities. For example, the probability of getting AT MOST 7 heads in 12 coin tosses is a cumulative probability equal to 0.806.)

• The probability of success for any individual student is 0.6.
• The number of trials is 3 (because we have 3 students).
• The number of successes is 2.

The calculator reports that the probability that two or fewer of these three students will graduate is 0.784.

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