Binomial Probability Calculator

Last updated: 2026-05-18

What Is the Binomial Distribution?

The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success. Tossing a coin 10 times and counting heads, testing 100 products and counting defects, or polling 50 voters — all follow the binomial model when each trial has only two outcomes and trials are independent.

Formulas

For $n$ trials, $k$ successes, and success probability $p$ per trial:

Combinations (number of ways to choose k successes from n trials):

C(n,k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}

Exact probability:

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

Cumulative probability (at most k successes):

P(X \leq k) = \sum_{i=0}^{k} \binom{n}{i} p^i (1-p)^{n-i}

At least k successes:

P(X \geq k) = 1 - P(X \leq k-1)

Mean and standard deviation:

\mu = np \qquad \sigma = \sqrt{np(1-p)}

Worked example: coin flips

Toss a fair coin ($p = 0.5$) 10 times. What is the probability of exactly 3 heads?

P(X = 3) = \binom{10}{3} \times 0.5^3 \times 0.5^7 = 120 \times \frac{1}{1024} \approx 0.117

About an 11.7% chance. The cumulative probability of 3 or fewer heads:

P(X \leq 3) = P(0) + P(1) + P(2) + P(3) = \frac{1 + 10 + 45 + 120}{1024} = \frac{176}{1024} \approx 17.2\%

And 3 or more heads: $P(X \geq 3) = 1 - P(X \leq 2) = 1 - 56/1024 \approx 94.5\%$ .

Normal approximation

For large $n$ , the binomial distribution approaches a bell curve. The normal approximation $X \approx N(\mu, \sigma^2)$ is reliable when:

np \geq 5 \quad \text{and} \quad n(1-p) \geq 5

A continuity correction adjusts for the shift from a discrete count to a continuous variable: $P(X = k) \approx P(k - 0.5 < Z < k + 0.5)$ , where $Z$ is the standard normal.

For $n = 10$, $p = 0.5$: $np = 5$ sits at the boundary, where the approximation is rough. At $n = 100$ or larger, the agreement is close.

Applications

Quality control: A factory produces bolts with a 2% defect rate. The probability of finding more than 2 defects in a batch of 50 is $P(X > 2) = 1 - P(X \leq 2)$ , with $n = 50$ and $p = 0.02$.

Medical testing: A drug has a 70% success rate. In a trial of 20 patients, the probability that at least 15 respond is $P(X \geq 15)$ , with $n = 20$ and $p = 0.7$.

Survey sampling: In a city where 40% support a policy, the probability that exactly 5 of 12 randomly selected voters support it is $P(X = 5)$ , with $n = 12$ and $p = 0.4$.

Frequently Asked Questions (FAQ)

What is the binomial probability formula?

P(X = k) = C(n,k) × p^k × (1−p)^(n−k), where C(n,k) = n! / (k!(n−k)!) is the number of ways to choose k successes from n trials. For example, the probability of exactly 3 heads in 10 fair coin flips: P = C(10,3) × 0.5³ × 0.5⁷ = 120 × (1/1024) ≈ 0.1172.

How do you calculate cumulative binomial probability?

P(X ≤ k) is the sum of P(X = 0) + P(X = 1) + … + P(X = k). For 3 or fewer heads in 10 flips: P(X ≤ 3) = P(0) + P(1) + P(2) + P(3) ≈ 0.172. P(X ≥ k) = 1 − P(X ≤ k−1), capturing the probability of k or more successes.

What is the mean and standard deviation of a binomial distribution?

Mean: μ = n × p. Standard deviation: σ = √(np(1−p)). For 10 trials with p = 0.5: μ = 5, σ = √(2.5) ≈ 1.58. These tell you the expected number of successes and how much the count typically varies around that expectation.

When can I approximate the binomial with a normal distribution?

The normal approximation is good when both np ≥ 5 and n(1−p) ≥ 5. Then X ≈ Normal(μ, σ²) with μ = np and σ² = np(1−p). Apply a continuity correction: P(X = k) ≈ P(k−0.5 < Z < k+0.5). For n = 10, p = 0.5, np = 5 — right at the boundary. More reliable for n ≥ 30.

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