Binomial Distribution Formula
Use the binomial distribution formula to calculate the likelihood an event will occur a specific number of times in a set number of opportunities. I’ll show you the binomial distribution formula to calculate these probabilities manually.
In this post, I’ll walk you through the formulas for how to find the probability, mean, and standard deviation of the binomial distribution and provide worked examples.
Note that this post focuses on the binomial distribution formulas and calculations. For more information about the distribution itself and how to use and graph it, please read Binomial Distribution: Uses & Calculator.
The binomial distribution formula for probabilities is the following:
where:
- n is the number of trials.
- x is the number of successes
- p is the probability of a success.
- (1–p) is the chance of failure.
Use this formula to calculate the binomial probability for X successes occurring in n trials. To find the correct solution, you must use the correct order of operations in the formula. If you need a refresher on that, read my post PEMDAS Explained: Order of Operations in Math.
nCx is the number of ways to obtain samples with the specified number of successes occurring within the set number of trials where the order of outcomes does not matter. Specifically, it’s the number of combinations without repetition. For more information, read my post about Finding Combinations.
The binomial distribution formula takes the number of combinations, multiplies that by the probability of success raised by the number of successes, and multiplies that by the probability of failures raised by the number of failures.
Let’s work through an example calculation to bring the formula to life!
Worked Example of Finding a Binomial Probability
We’ll use the binomial distribution formula to calculate the chances of rolling exactly three sixes in ten die rolls for this example. Here are the values to enter into the formula:
- n = 10
- x = 3
- p = 0.1667
- (1–p) = 0.8333
For the number of combinations, we have:
Now, let’s enter our values into the binomial distribution formula.
This calculation finds that the binomial probability of rolling three 6s in 10 rolls is 0.1540.
If you need to calculate a cumulative probability for a binomial random variable, calculate the likelihood for each individual outcome and then sum them for all outcomes of interest.
For example, if you want to calculate the probability of ≥ 3 sixes in 10 rolls, calculate the likelihoods for three sixes, four sixes, etc., on up to ten sixes. Then sum that set of binomial probabilities.
Read on to learn about the formulas to calculate the mean and standard deviation of the binomial distribution!
Use my Binomial Distribution Calculator to find the probability of a specific number of events occurring in a fixed number of trials.
Expected Value of Binomial Distribution
The expected value of the binomial distribution is its mean.
The binomial distribution formula for the expected value is the following:
n * p
Multiply the number of trials (n) by the success probability (p). This value represents the average or expected number of successes.
For example, we roll the die ten times, and the probability of rolling a six is 0.1667. Let’s enter these values into the formula
10 * 0.1667
The mean for this binomial distribution is 1.667. On average, we’d expect to roll that many sixes in ten rolls. Of course, the actual counts of successes will always be either zero or a positive integer.
This mean is the expected value for a binomial distribution. Learn more about Expected Values: Definition, Formula & Finding.
Standard Deviation of the Binomial Distribution
The binomial distribution formula for the standard deviation is the following:
As before, n and p are the number of trials and success probability, respectively. (1 – p) is the likelihood of failure.
Notice that the standard deviation of the binomial distribution is at its maximum when the probabilities for success and failure are both 0.5. As those probabilities move away from 0.5 in opposite directions, it decreases. Additionally, it increases as the number of trials (n) increase.
For our die example we have n = 10 rolls, a success probability of p = 0.1667, and a failure probability of (1 – p) = 0.833. Let’s enter these values into the formula.
10 * 0.1667 * 0.8333 = 1.3891. That’s the variance, which uses squared units.
To find the standard deviation of the binomial distribution, we need to take the square root of the variance.
The standard deviation represents the variability of the probabilities around the mean of the binomial distribution. Learn more about the Standard Deviation.
By using the formula for the binomial distribution, it is easy to calculate its probabilities, means, and standard deviations.
Dear Jim,
I would like to have a question regarding the distribution of outcomes of multiple binomial distributions.
I have a sample of 50 subjects, where every subject completes a task with two possible outcomes (left or right hand use, with 50% probability) 30 times. On an individual level, this leads to a binomial distribution of outcomes for each subject, where a z-score can be calculated for each subject to see how far the hand uses are from the expected 50% (15 left and right hand use) level. Based on the z-scores, we deem those with larger values than 1.96 (›1.96) as right-handed, those with lower values than -1.96 (‹-1.96) as left-handed and those with values in between (-1.96‹‹1.96) as ambilateral.
Question: what is the expected ratio of z-scores (or distribution of z-scores) in my sample of 50 subjects? Is it the same as at the individual level with 95% chance of being ambilateral and 2.5-2.5% as being left or right-handed?
I have a probability question I have been trying to figure out.
Here is the scenario:
A group (12) of golfers gather each Saturday to play in a match. Each player is trying to gain points based on their individual handicap. For example, my handicap is 34 and if I get 34 pts I finish Even for the day. If I get 32 pts, I finish -2; 36 pts, +2 and so on.
At the end of the match players are randomly drawn as teammates from a hat. The team with the highest net score wins the match. Let’s assume for this match, teams will be drawn out in 2s. Let’s also assume that the maximum + or – for each player is 5. For instance, I may be drawn with player #7 who is +4 for the day and I am -1 for the day. Our team score is +3.
What is the probability that I (my team) will win? Do my odds change based on my score?
What if there are 20 players? Or, if teams are divided into 4s?
Thanks for providing some help and insight!