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, as well as to learn the binomial probability formula. Find out what is binomial distribution, and discover how binomial experiments are used in various settings.

## What is the binomial probability?

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 (exactly 3); you also are able to 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 in order to effectively use the binomial distribution calculator, the events you analyze must be **independent**. It means that all the trials in your example are supposed to be mutually exclusive.

The success in the first trial 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 game a dice 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 time. Such questions may be addressed using a related statistical tool called the negative binomial distribution. Make sure to read about the differences between this distribution and the negative binomial distribution.

## Binomial probability formula

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

`P(X=r) = nCr * p^r * (1-p)^(n-r)`

where:

**n**is the total number of events;**r**is the number of required successes;**p**is the probability of one success;**nCr**is the number of combinations (so-called "n choose r");**P(X=r)**is the 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 exactly 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, exactly four and exactly five successes and sum all of these values together.

You can use the SMp(x) probability distribution to simulate many other distributions including the binomial one. Make sure to give it a try!

## How to use the binomial distribution calculator: an example

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

- Determine the number of events.
**n**is equal to 5, as we roll 5 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. - 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 :)

## Binomial probability distribution experiments

The binomial distribution turns out be very practical in **experimental settings**. The output of such a random experiment, however, 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 a dice or tossing a coin, it may be employed in somehow less obvious 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`

. You can convert percentage to fraction, or the other way around, by using this calculator.

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 deficient 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.

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

## Mean and variance of binomial distribution

One of the most interesting features of binomial distributions is that they represents 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 the addition properties of random variables 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 a square root of variance, `σ`

. Keep it mind that the standard deviation calculated from your sample (the observations you actually gather) may differ from the standard deviation of the entire population. 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 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 10 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 trails we expect to get anywher 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.

## Other considerations

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 very large, 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 it mind that the binomial distribution formula describes a **discrete distribution**. The possible outcomes of all the trails 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 in 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 trails, 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 problem of the dice game again, but this time you need 3 or more successes to win it. How about the chances of getting exactly 4?