SigmaCalc

Binomial Probability Calculator

The math behind coin flips, quality-control sampling, and any process with a fixed 'success or not' probability per trial.

Number of trials (n)

Number of successes (k)

Success probability (0–1)

P(X = k)

0.205

P(X = k) = C(n,k) × pᵏ × (1−p)ⁿ⁻ᵏ

What you entered

Number of trials (n): 10Number of successes (k): 4Success probability (0–1): 0.5

1
Number of ways to choose k successes
C(10, 4)= 210
2
Probability of that exact sequence
0.5^4 × (1 − 0.5)^6= 0.001
3
P(X = k)
210 × 0.001= 0.2051

Result

P(X = k): 0.205

Out of 10 independent trials with a 50% success probability, there's a 20.51% chance of exactly 4 successes.

Binomial probability of exactly k successes in n trials

Each row is the chance of exactly k successes in n independent trials, each with success probability p.

Trials n	Successes k	p	Probability
10	5	0.5	0.2461
10	3	0.5	0.1172
20	10	0.5	0.1762
10	2	0.2	0.3020
6	6	0.5	0.0156
12	4	0.25	0.1936

The first row is the most instructive: exactly 5 heads in 10 fair coin tosses is only 24.61% likely, even though 5 is the single most probable outcome. People routinely overestimate this, because the most likely individual result still leaves three quarters of the probability spread across every other count. Row three shows the same effect more strongly - exactly 10 heads in 20 tosses is just 17.62%. Row five is six heads in six tosses, at 1 in 64. The formula requires independent trials with a constant p; sampling without replacement breaks that and needs the hypergeometric distribution instead.

When the binomial distribution applies

It requires a fixed number of independent trials, each with only two possible outcomes, and the same success probability every time — think repeated coin flips, not something where the odds change as you go.

Beyond exactly k

This calculator gives P(X = k), the probability of exactly that many successes. To find P(X ≤ k) or P(X ≥ k), you'd sum this formula across the relevant range of k values.

Frequently asked questions

A factory's defect rate is 2%. In a batch of 50, what's the probability of exactly 3 defective items?

n=50, p=0.02, k=3. P(X=3) = C(50,3) × 0.02³ × 0.98⁴⁷ = 19,600 × 0.000008 × 0.384 = 0.0607 = 6.1%. The most likely outcomes are 0 defects (36.4%) and 1 defect (37.2%).

How is the binomial distribution different from the Poisson distribution?

Binomial: fixed number of trials, each with the same probability (flip a coin 10 times). Poisson: counts events in a fixed interval at a known average rate (how many calls per hour). When n is large and p is small, binomial approximates Poisson with λ = np.

What are the conditions for using the binomial distribution?

Four conditions: (1) Fixed number of trials n. (2) Each trial has exactly two outcomes (success/failure). (3) Probability p is the same for every trial. (4) Trials are independent. If any condition fails, the binomial model doesn't apply.