🎲 Enter Parameters & Target Successes
Quick Examples
Total number of trials (1 to 500)
n must be a whole number from 1 to 500.
[0,1]
Probability of success on each trial
p must be between 0 and 1 (exclusive).
For range: enter lower bound here
k must be 0 to n (whole number).
P(X = k)
0.0000
Binomial(n, p) probability
📐 Step-by-Step Working
📋 Full Probability Mass Function — P(X = k) for all k
| k | P(X=k) | P(X≤k) | P(X≥k) | Distribution |
|---|
⚠️ Note: This calculator uses exact binomial arithmetic. For very large n (>400), minor floating-point rounding may occur. For n>500 or for very small p with large n, consider using a normal approximation with continuity correction (valid when np ≥ 10 and n(1−p) ≥ 10).
Sources & Methodology
Khan Academy — Binomial Random Variables
Primary reference for the binomial PMF formula P(X=k) = C(n,k)×pk×(1−p)n−k, the four conditions for binomial experiments, and worked examples verifying the calculation logic used in this calculator.
NIST/SEMATECH e-Handbook — Binomial Distribution
Authoritative source for mean = np, variance = np(1−p), standard deviation = √(np(1−p)), and the normal approximation condition np≥5 and n(1−p)≥5.
Formulas verified (manual check: n=10, p=0.5, k=3):
PMF: P(X=k) = C(n,k) × pk × (1−p)n−k
C(10,3) = 120 | 0.53=0.125 | 0.57=0.0078125 | P(X=3) = 120×0.125×0.0078125 = 0.1172 ✓
Mean=np=5 | Variance=np(1−p)=2.5 | SD=1.5811 ✓
PMF: P(X=k) = C(n,k) × pk × (1−p)n−k
C(10,3) = 120 | 0.53=0.125 | 0.57=0.0078125 | P(X=3) = 120×0.125×0.0078125 = 0.1172 ✓
Mean=np=5 | Variance=np(1−p)=2.5 | SD=1.5811 ✓
Binomial Distribution — Formula, Worked Examples & When to Use It
You’re already using binomial distribution without realizing it any time you ask “what’s the chance of getting 7 heads in 10 coin flips?” or “if 30% of customers churn, what’s the probability exactly 4 of my next 20 customers leave?” Those are both binomial probability problems. The formula is the same in both cases — it just needs n (number of trials), p (probability of success), and k (target number of successes).
The Four Conditions: When You Can Use Binomial Distribution
The binomial distribution only applies when all four of these conditions hold. Missing any one of them means you need a different model.
- Fixed number of trials (n). You know in advance how many times you’re repeating the experiment. 10 coin flips, 20 products inspected, 50 customers surveyed.
- Two outcomes per trial. Each trial must be classifiable as success or failure. Heads/tails, defective/not defective, yes/no, pass/fail.
- Constant probability (p). The probability of success must be the same on every trial. A fair coin is always 0.5. A factory defect rate of 5% applies to every item.
- Independent trials. The outcome of one trial doesn’t change the probability for the next. Coin flips are independent. Sampling without replacement from a small population is not — use the hypergeometric distribution instead.
The Formula — Worked Through a Real Example
A free throw shooter makes 75% of shots. She takes 10 free throws. What’s the probability she makes exactly 7?
🧮 Binomial PMF Formula (Verified)
P(X = k) = C(n, k) × pk × (1 − p)n − k
C(n, k) = n! / (k! × (n−k)!) (binomial coefficient)
Mean = n × p | Variance = n × p × (1−p) | SD = √(n×p×(1−p))
Worked: n=10, p=0.75, k=7
C(10,7) = 10! / (7! × 3!) = 120
0.757 = 0.1335 0.253 = 0.0156
P(X=7) = 120 × 0.1335 × 0.0156 = 0.2503 (about 25%)
Mean = 7.5 shots expected | SD = 1.369 | Source: NIST e-Handbook + Khan Academy ✓
C(10,7) = 10! / (7! × 3!) = 120
0.757 = 0.1335 0.253 = 0.0156
P(X=7) = 120 × 0.1335 × 0.0156 = 0.2503 (about 25%)
Mean = 7.5 shots expected | SD = 1.369 | Source: NIST e-Handbook + Khan Academy ✓
Understanding Exact vs Cumulative Probabilities
P(X = 7) is the probability of exactly 7 makes. P(X ≤ 7) is the probability of 7 or fewer — you add up P(X=0) through P(X=7). P(X ≥ 7) is the probability of 7 or more, which equals 1 − P(X ≤ 6). Most real-world questions are cumulative: “what’s the chance of catching at least 2 defects?” or “what’s the probability of no more than 3 failures?”
With n=10, p=0.75 (continuing the example above): P(X=7) = 0.2503 but P(X ≥ 7) = P(7)+P(8)+P(9)+P(10) = 0.2503+0.2816+0.1877+0.0563 = 0.7759. A 78% chance she makes 7 or more from 10 attempts. That’s the number that matters if she needs to make at least 7 to win a competition.
Common Mistakes That Give Wrong Answers
The PMF calculation is straightforward but there are a few places where it’s easy to go wrong:
| Mistake | Example of the Error | Correct Approach |
|---|---|---|
| Using p instead of 1−p for failures | P(X=7) = C(10,7)×0.757×0.753 | The exponent on the failure side is (1−p), not p. Use 0.253. |
| Confusing P(X=k) and P(X≤k) | Answering “at most 3” with just P(X=3) | P(X≤3) = sum P(X=0)+P(X=1)+P(X=2)+P(X=3) |
| Using ≥ when the problem says > | P(X > 5) answered as P(X ≥ 5) | P(X > 5) = 1 − P(X ≤ 5). P(X ≥ 5) = 1 − P(X ≤ 4). Off by one k matters. |
| Applying binomial when sampling without replacement | Drawing 5 cards from a 52-card deck with binomial | When sampling a meaningful fraction of a small population, use hypergeometric distribution. |
| Forgetting to check independence | Using binomial for customer purchases from same household | If one outcome influences another, trials aren’t independent. Binomial doesn’t apply. |
💡 Normal approximation rule of thumb: The binomial distribution starts looking bell-shaped and symmetric when np ≥ 10 AND n(1−p) ≥ 10. At that point you can use the normal distribution with mean = np and SD = √(np(1−p)). But for small n or extreme p (very close to 0 or 1), always use the exact binomial. An n=10, p=0.05 scenario has np=0.5 — way below the threshold. The exact binomial is non-negotiable there.
Frequently Asked Questions
P(X = k) = C(n,k) × pk × (1−p)n−k. C(n,k) = n!/(k!(n−k)!) counts the number of ways to get k successes from n trials. pk is the probability of k successes. (1−p)n−k is the probability of the remaining (n−k) failures. Multiply them together and you get the exact probability of exactly k successes. Example: n=5, p=0.4, k=2 → C(5,2)×0.42×0.63 = 10×0.16×0.216 = 0.3456.
P(X ≥ k) = 1 − P(X ≤ k−1). Sum the PMF from 0 to k−1, then subtract from 1. Example: n=6, p=0.5, P(at least 4 heads) = 1 − P(X≤3) = 1 − [P(0)+P(1)+P(2)+P(3)] = 1 − 0.6563 = 0.3438. Using this calculator: select “P(X ≥ k)” from the dropdown and it handles the subtraction automatically.
Mean = n × p. This is the expected number of successes over many repetitions. Standard deviation = √(n×p×(1−p)). For n=20 coin flips: mean = 10, SD = 2.236 — so you’d typically expect between 7.8 and 12.2 heads (within one SD). The mean tells you the center; the SD tells you the spread. Larger n or p closer to 0.5 gives larger SD.
Fixed n (you know how many trials before starting), binary outcomes (success or failure only), constant probability p (same on every trial), independent trials (one trial doesn’t affect the next). All four must hold. If you’re sampling from a small population without replacement, trials aren’t independent — use hypergeometric. If p changes between trials, binomial doesn’t apply.
C(n,k) = n! / (k! × (n−k)!). It counts how many ways you can arrange k successes in n trials. With 5 coin flips, how many ways can exactly 3 be heads? C(5,3) = 5!/(3!2!) = 10. There are 10 distinct orderings of HHHTTs (ignoring which toss is which). That count is the binomial coefficient. C(n,0) = C(n,n) = 1 always. C(n,1) = n always. This is the same as combinations in combinatorics.
When np ≥ 10 AND n(1−p) ≥ 10, the binomial looks close to normal with mean=np and SD=√(np(1−p)). Example: n=100, p=0.3 gives np=30 and n(1−p)=70 — both well above 10, normal approximation is reliable. With n=20, p=0.05: np=1, far below 10 — the distribution is heavily right-skewed and normal approximation fails completely. Always use exact binomial when the rule is violated.
The PMF (Probability Mass Function) table shows P(X=k) for every possible value of k from 0 to n. It gives you the full shape of the distribution at a glance. The peak of the PMF table is at (or near) the mean np. If p < 0.5, the distribution is right-skewed — more weight on low values. If p > 0.5, it’s left-skewed. If p = 0.5, it’s symmetric. The table also shows cumulative P(X≤k) and P(X≥k) so you can answer any cumulative question without recalculating.
P(X = k) is the exact (point) probability of k successes. P(X < k) = P(X ≤ k−1) is the probability of fewer than k successes — strictly less than, not including k itself. Since binomial is discrete, P(X < k) ≠ P(X ≤ k). For example, P(X < 5) = P(X ≤ 4) = sum P(0) through P(4). P(X ≤ 5) = sum P(0) through P(5). That extra P(5) can be significant. This off-by-one error is one of the most common mistakes in binomial problems.
Technically yes, but trivially. If p = 1, every trial is a success and P(X=n) = 1. If p = 0, no trial is a success and P(X=0) = 1. These are degenerate cases. In practice, you wouldn’t use the binomial formula — the answer is obvious. This calculator requires p strictly between 0 and 1 to avoid degenerate inputs.
The mode is the k with the highest P(X=k). When (n+1)×p is not an integer, mode = floor((n+1)×p). When it is an integer, there are two modes. For n=10, p=0.5: (11)(0.5)=5.5, floor=5, so mode=5. For n=10, p=0.3: (11)(0.3)=3.3, floor=3, mode=3. The mode is always near the mean np. You can also just read it from the PMF table — the row with the highest P(X=k) value.
Quality control uses binomial to set acceptance sampling rules. If a factory produces items with a 3% defect rate and inspectors sample 50 items, P(X ≥ 3 defects) tells you the probability of catching or rejecting the batch. Acceptance sampling plans (like MIL-STD-1916) use binomial to balance the risk of accepting bad batches vs rejecting good ones. Binomial is also used for attribute control charts (p-charts) tracking defect rates over time.
Use Poisson when n is very large and p is very small, so np is moderate (a few to a few hundred). Rule of thumb: if n > 20 and p < 0.05, Poisson with λ = np is a good approximation to binomial. Example: a factory produces 10,000 items with a 0.1% defect rate. n=10,000, p=0.001, np=10. Poisson(10) is much easier to compute and essentially identical to Binomial(10000, 0.001). For larger p or smaller n, stick with exact binomial.
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