Binomial Distribution Calculator

Find exact and cumulative binomial probabilities, mean, variance, and a full distribution table for any n, k, and p.

🎲 Binomial Distribution Calculator
Number of Trials (n)10
1100
Number of Successes (k)3
0100
Success Probability (p)50%
%
0%100%
Number of Trials (n)8
130
Success Probability (p)50%
%
0%100%
P(X = k) Exact
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P(X ≤ k) Cumulative
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P(X ≥ k) Upper Tail
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Mean (np)
—
Variance np(1-p)
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Std Deviation
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Mean (np)
—
Std Deviation
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🎲 What is the Binomial Distribution?

The binomial distribution is one of the most fundamental probability distributions in statistics. It describes the number of successes in a fixed number of independent trials, where each trial has exactly two possible outcomes (success or failure) and the probability of success is the same on every trial.

Real-world scenarios that follow the binomial distribution are everywhere. A quality control engineer inspecting 50 parts for defects, where each part has a 2% defect rate, is working with a binomial distribution. A basketball player taking 15 free throws with a 70% success rate produces binomial outcomes. A medical researcher counting how many of 100 patients respond to a drug with a 60% response rate is also modelling a binomial situation. Any repeated, independent yes-or-no process with a fixed success rate is binomial.

A common misconception is confusing the binomial distribution with the normal distribution. The binomial distribution is discrete (whole number outcomes only) and is defined on the integers 0, 1, 2, ..., n. The normal distribution is continuous. However, when n is large enough and p is not too close to 0 or 1 (specifically when both np and n(1-p) exceed 5), the binomial distribution is well approximated by a normal distribution with the same mean and variance. This is a powerful result that underlies much of statistical inference.

The binomial distribution is also distinct from the Poisson distribution (which models rare events over time) and the geometric distribution (which models the number of trials until the first success). Understanding which distribution applies requires checking the four conditions: fixed n, independent trials, binary outcomes, and constant p across all trials.

📐 Formula

P(X = k)  =  C(n, k) × pk × (1 − p)n − k
n = number of independent trials
k = number of successes (0, 1, 2, ..., n)
p = probability of success on a single trial (0 to 1)
C(n, k) = n! ÷ (k! × (n − k)!)  [the binomial coefficient]
Mean = np
Variance = np(1 − p)
Std Deviation = √(np(1 − p))
Example: n = 10, k = 3, p = 0.5: C(10,3) × 0.53 × 0.57 = 120 × 0.125 × 0.0078 = 11.72%

📖 How to Use This Calculator

Steps

1
Choose your mode - Select Calculate Probability to find P(X=k) and cumulative probabilities for a specific k, or select Distribution Table to generate the full PMF and CDF for all k from 0 to n.
2
Enter number of trials - Set n to the total number of independent trials (e.g., 20 coin flips, 50 quality inspections). Probability mode supports up to n = 1000; Table mode supports up to n = 30.
3
Enter k and probability - In Probability mode, set k to the exact number of successes you want to evaluate, and p (%) to the per-trial success rate. In Table mode, only n and p are needed.
4
Read the results - The calculator shows exact probability P(X=k), cumulative P(X ≤ k), upper tail P(X ≥ k), mean, variance, and standard deviation. The Distribution Table lists every outcome with its probability and CDF value.

💡 Example Calculations

Example 1 - Fair Coin Flipped 10 Times

What is the probability of getting exactly 3 heads in 10 fair coin flips?

1
Set n = 10, k = 3, p = 50%.
2
Binomial coefficient: C(10, 3) = 10! / (3! x 7!) = 120.
3
P(X = 3) = 120 x 0.53 x 0.57 = 120 x 0.125 x 0.0078125 = 11.72%.
4
Mean = 10 x 0.5 = 5 heads. Std Dev = sqrt(10 x 0.5 x 0.5) = 1.58.
P(X = 3) = 11.72%, P(X ≤ 3) = 17.19%, Mean = 5
Try this example →

Example 2 - Manufacturing Defect Rate

A factory has a 5% defect rate. What is the probability of finding exactly 2 defects in a batch of 20?

1
Set n = 20, k = 2, p = 5%.
2
C(20, 2) = 20! / (2! x 18!) = 190.
3
P(X = 2) = 190 x 0.052 x 0.9518 = 190 x 0.0025 x 0.3972 = 18.87%.
4
Mean = 20 x 0.05 = 1 defect. P(X ≤ 2) = 92.45% (most batches have at most 2 defects).
P(X = 2) = 18.87%, P(X ≤ 2) = 92.45%, Mean = 1.0
Try this example →

Example 3 - Drug Trial Response Rate

A drug has a 60% response rate. In a trial of 15 patients, what is the probability that 10 or more respond?

1
Set n = 15, k = 10, p = 60%. We need P(X ≥ 10) from the upper tail result.
2
P(X ≥ 10) = P(X=10) + P(X=11) + ... + P(X=15), summed from the upper tail.
3
Mean = 15 x 0.6 = 9 patients. Std Dev = sqrt(15 x 0.6 x 0.4) = 1.897.
4
P(X ≥ 10) = 40.32% (the calculator sums P(X=10) through P(X=15) automatically).
P(X ≥ 10) = 40.32%, Mean = 9, Std Dev = 1.897
Try this example →

❓ Frequently Asked Questions

What is the binomial distribution formula?+
The formula is P(X = k) = C(n,k) x p^k x (1-p)^(n-k), where n is the number of trials, k is the number of successes, p is the success probability per trial, and C(n,k) = n! / (k!(n-k)!) is the binomial coefficient. This gives the probability of getting exactly k successes in n independent trials.
What are the four conditions for the binomial distribution?+
The four conditions are: (1) a fixed number of trials n; (2) each trial is independent (the outcome of one trial does not affect others); (3) each trial has only two possible outcomes (success and failure); and (4) the probability of success p is constant across all trials. If any condition fails, the binomial model does not apply.
What is the mean and standard deviation of the binomial distribution?+
The mean is mu = np. The variance is sigma^2 = np(1-p). The standard deviation is sigma = sqrt(np(1-p)). For example, with n = 50 and p = 0.4, mean = 20, variance = 12, standard deviation = 3.464. These formulas hold for any valid n and p.
What is the difference between P(X=k), P(X≤k), and P(X≥k)?+
P(X=k) is the exact probability of getting exactly k successes (the PMF value). P(X≤k) is the cumulative probability of getting k or fewer successes (the CDF value). P(X≥k) is the upper tail probability of getting k or more successes. Note that P(X≤k) + P(X≥k) - P(X=k) = 1, since P(X=k) is counted in both cumulative terms.
When can I use the normal approximation to the binomial?+
The normal approximation N(np, np(1-p)) is reliable when both np > 5 and n(1-p) > 5. For example, with n = 100 and p = 0.3, np = 30 and n(1-p) = 70, both well above 5. For better accuracy, apply a continuity correction: P(X = k) is approximated by P(k - 0.5 < Z < k + 0.5) where Z is the standard normal variable.
What is the difference between the binomial and Poisson distributions?+
The binomial distribution counts successes in a fixed number of trials n with constant probability p. The Poisson distribution counts events in a fixed time or space interval, with no upper bound on the count. As n becomes large and p becomes small while np = lambda stays constant, the binomial converges to the Poisson distribution with parameter lambda = np.
How do I calculate cumulative binomial probability by hand?+
P(X ≤ k) is the sum of P(X = 0) + P(X = 1) + ... + P(X = k), where each term uses C(n,i) x p^i x (1-p)^(n-i). For large n this becomes tedious, which is why calculators and statistical tables are used. The Distribution Table mode above shows all values instantly.
Can the binomial distribution be used for sampling without replacement?+
Technically, sampling without replacement violates the independence requirement. However, if the population is large relative to the sample size (population N >= 20 x sample size n), the binomial is a good approximation. When independence is truly violated, the hypergeometric distribution is the correct model to use.
What is the mode of the binomial distribution?+
The mode is the most probable value of X. For a binomial(n, p) distribution, the mode is floor((n+1)p) when (n+1)p is not an integer. When (n+1)p is an integer, both (n+1)p and (n+1)p - 1 are modes. The Distribution Table mode above makes it easy to identify the maximum probability row visually.
How does a binomial distribution change shape for different p values?+
For p = 0.5 the distribution is perfectly symmetric around np. For p < 0.5 the distribution is right-skewed, with most probability weight at low k values. For p > 0.5 it is left-skewed. As n increases, the distribution becomes more symmetric and increasingly resembles a normal curve, regardless of p, by the Central Limit Theorem.
What is the complement rule for binomial probability?+
The complement rule states P(X ≥ k) = 1 - P(X ≤ k-1). This is useful when computing the probability of at least k successes is easier via the complement. For example, P(at least 1 success) = 1 - P(X = 0) = 1 - (1-p)^n, which is far simpler than summing P(X=1) through P(X=n).
Tags: Binomial Distribution Calculator Binomial Probability P(X=k) Cumulative Binomial Probability Calculator Formula Free Online

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What is the binomial distribution formula?

P(X = k) = C(n,k) x p^k x (1-p)^(n-k), where n is the number of trials, k is the target number of successes, p is the probability of success on each trial, and C(n,k) = n! / (k! x (n-k)!) is the binomial coefficient.

What is the mean of the binomial distribution?

The mean (expected value) is mu = np. For example, with n = 20 trials and p = 0.3, the expected number of successes is 20 x 0.3 = 6.

What is the variance of the binomial distribution?

The variance is sigma^2 = np(1-p). The standard deviation is sigma = sqrt(np(1-p)). For n = 20 and p = 0.3, variance = 4.2 and standard deviation = 2.049.

When can I use the normal approximation to the binomial?

The normal approximation is reliable when both np > 5 and n(1-p) > 5. Apply a continuity correction for better accuracy.

What is the difference between the PMF and CDF?

The PMF gives P(X = k). The CDF gives P(X <= k) = sum of P(X = 0) through P(X = k).

What does P(X >= k) mean?

P(X >= k) is the upper tail probability — the chance of getting k or more successes. It equals 1 minus P(X <= k-1), i.e., one minus the cumulative probability up to k-1. For example, P(X >= 3) = 1 - P(X <= 2) = 1 - [P(0) + P(1) + P(2)].

How do I calculate binomial probability for large n?

For large n use log-probability: log P(X=k) = log C(n,k) + k log p + (n-k) log(1-p). This calculator handles n up to 1000.

What are the conditions for the binomial distribution?

Four conditions: fixed n, independent trials, binary outcomes (success/failure), and constant probability p across all trials.

📌 Quick Tips

💡When p = 0.5 the distribution is perfectly symmetric, so P(X = k) = P(X = n-k).
💡For large n and small p (n > 20, p < 0.05), the Poisson distribution with lambda = np gives a good approximation.
💡The mean of a binomial distribution is always np, regardless of how the probability is distributed across trials.
💡For n > 30 and np > 5 and n(1-p) > 5, the normal distribution N(np, np(1-p)) approximates the binomial well.
💡Use the Distribution Table mode to see all probabilities at once and identify the most likely outcomes.