Binomial Distribution Calculator
Our free descriptive & distributions calculator solves binomial distribution problems. Get worked examples, visual aids, and downloadable results.
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer
Binomial Distribution Calculator
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Formula: P(X = k) = C(n,k) x p^k x (1-p)^(n-k)
Worked example — P(X=2) = 18.87% | P(X<=2) = 92.45% | Mean = 1.0 defect
Formula
P(X = k) = C(n,k) x p^k x (1-p)^(n-k)
The binomial probability formula calculates the chance of getting exactly k successes in n independent trials, each with success probability p. C(n,k) is the binomial coefficient (number of ways to choose k items from n), p^k is the probability of k successes, and (1-p)^(n-k) is the probability of the remaining failures.
Worked Examples
Example 1: Quality Control Inspection
Problem:A factory has a 5% defect rate. In a random sample of 20 items, what is the probability of finding exactly 2 defective items?
Solution:n = 20 trials, p = 0.05, k = 2 P(X = 2) = C(20,2) x 0.05^2 x 0.95^18 C(20,2) = 20! / (2! x 18!) = 190 P(X = 2) = 190 x 0.0025 x 0.3972 = 0.1887 Mean: 20 x 0.05 = 1.0 defects expected Std Dev: sqrt(20 x 0.05 x 0.95) = 0.9747 P(X <= 2): 0.9245 (92.45% chance of 2 or fewer defects)
Result:P(X=2) = 18.87% | P(X<=2) = 92.45% | Mean = 1.0 defect
Example 2: Medical Trial Success Rate
Problem:A drug has a 70% effectiveness rate. In a trial of 15 patients, what is the probability that at least 12 respond positively?
Solution:n = 15, p = 0.70, find P(X >= 12) P(X >= 12) = P(X=12) + P(X=13) + P(X=14) + P(X=15) P(X=12) = C(15,12) x 0.70^12 x 0.30^3 = 0.1700 P(X=13) = C(15,13) x 0.70^13 x 0.30^2 = 0.0916 P(X=14) = C(15,14) x 0.70^14 x 0.30^1 = 0.0305 P(X=15) = C(15,15) x 0.70^15 x 0.30^0 = 0.0047 P(X >= 12) = 0.2968
Result:P(X>=12) = 29.68% | Mean = 10.5 patients | Std Dev = 1.77
Frequently Asked Questions
What is the binomial distribution and when is it used?
The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. It is used when an experiment has exactly two possible outcomes (success or failure), the number of trials is fixed in advance, each trial is independent, and the probability of success remains constant. Common applications include quality control (number of defective items in a batch), medical trials (number of patients responding to treatment), survey analysis (number of respondents choosing a particular option), genetics (inheritance patterns), and marketing (conversion rates). The distribution is characterized by two parameters: n (number of trials) and p (probability of success).
How do you calculate binomial probability using the formula?
The binomial probability formula is P(X = k) = C(n,k) x p^k x (1-p)^(n-k), where n is the total number of trials, k is the desired number of successes, p is the probability of success on a single trial, and C(n,k) is the binomial coefficient calculated as n! / (k!(n-k)!). The binomial coefficient counts the number of ways to choose k successes from n trials. The term p^k represents the probability of getting exactly k successes, and (1-p)^(n-k) represents the probability of getting the remaining (n-k) failures. For example, the probability of getting exactly 3 heads in 10 coin flips is C(10,3) x 0.5^3 x 0.5^7 = 120 x 0.000977 = 0.1172 or about 11.72%.
What is the difference between PMF and CDF in binomial distribution?
The PMF (Probability Mass Function) gives the probability of getting exactly k successes, written as P(X = k). The CDF (Cumulative Distribution Function) gives the probability of getting k or fewer successes, written as P(X <= k), and is calculated by summing the PMF values from 0 to k. The CDF is particularly useful for answering questions like 'what is the probability of getting at most 3 successes?' or 'what is the probability of getting at least 4 successes?' (which equals 1 minus the CDF at 3). Understanding the distinction between PMF and CDF is essential for correctly interpreting statistical results and making decisions based on probability thresholds in hypothesis testing and quality control applications.
How do mean and standard deviation relate to the binomial distribution?
For a binomial distribution with n trials and success probability p, the mean (expected value) is simply mu = n x p, representing the average number of successes you would expect. The variance is sigma-squared = n x p x (1-p), and the standard deviation is sigma = sqrt(n x p x (1-p)). These formulas reveal important properties: the mean increases linearly with n, and the variance is maximized when p = 0.5 (maximum uncertainty). The standard deviation tells you how spread out the distribution is around the mean. For large n, approximately 68% of outcomes fall within one standard deviation of the mean, and 95% within two standard deviations, following the normal approximation to the binomial distribution.
When can you approximate the binomial distribution with other distributions?
The binomial distribution can be approximated by other distributions under certain conditions, which simplifies calculations for large sample sizes. The normal approximation is valid when both np >= 5 and n(1-p) >= 5, using a normal distribution with mean np and standard deviation sqrt(np(1-p)). A continuity correction of plus or minus 0.5 improves accuracy. The Poisson approximation is appropriate when n is large (typically over 20) and p is small (typically under 0.05), using lambda = np as the Poisson parameter. This is commonly used in rare event modeling. These approximations were historically important for hand calculations but remain conceptually valuable for understanding the relationships between probability distributions and for quick mental estimates in statistical analysis.
References
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer · Editorial policy
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