Binomial Probability Calculator — P(X = k) & Cumulative (Free)
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A binomial probability calculator finds the chance of a given number of successes in n independent trials. The formula is P(X = k) = C(n, k) × pk × (1 − p)n − k. For 3 successes in 10 trials with p = 0.3: 120 × 0.027 × 0.0823543 = 0.2668, so about a 27% chance.
Which probability?
| Method | Probability | Error |
|---|---|---|
| Exact binomial | 0.2668 | — |
| Normal with continuity correction | 0.2699 | 0.003102 |
| Normal without correction | 0 | −0.266828 |
Show steps
- Check the binomial setting: a fixed number of trials n = 10, each a success or a failure, independent, with the same success probability p = 0.3. Failure probability: q = 1 − p = 0.7.
- Formula for one value: P(X = x) = C(n, x) × p^x × (1 − p)^(n − x), where C(n, x) = n! / (x!(n − x)!) counts the ways to place x successes among n trials.
- For x = 3: C(10, 3) = 120, p^3 = 0.3^3 = 0.027, q^7 = 0.7^7 = 0.082354. Product: P(X = 3) = 0.2668.
- Mean and spread: μ = np = 3, σ² = np(1 − p) = 2.1, σ = √σ² = 1.4491.
Distribution table
| x | P(X = x) | P(X ≤ x) | P(X ≥ x) |
|---|---|---|---|
| 0 | 0.0282 | 0.0282 | 1 |
| 1 | 0.1211 | 0.1493 | 0.9718 |
| 2 | 0.2335 | 0.3828 | 0.8507 |
| 3 | 0.2668 | 0.6496 | 0.6172 |
| 4 | 0.2001 | 0.8497 | 0.3504 |
| 5 | 0.1029 | 0.9527 | 0.1503 |
| 6 | 0.0368 | 0.9894 | 0.0473 |
| 7 | 0.0090 | 0.9984 | 0.0106 |
| 8 | 0.0014 | 0.9999 | 0.0016 |
| 9 | 0.0001 | ≈ 1.0000 | 0.0001 |
| 10 | 5.905 × 10^−6 | ≈ 1.0000 | 5.905 × 10^−6 |
How to use the binomial probability calculator
- Enter the number of trials n, the probability of success on one trial p (as 0.3, 30% or 1/6) and the number of successes k.
- Choose the probability you need: exactly, at most, fewer than, at least, more than or between two values.
- Read the answer, then all five single and cumulative probabilities, the mean, the variance and the standard deviation. Open Show steps for the working and Distribution table for every value of X.
The bar chart highlights the outcomes your question counts, and the normal approximation check compares the exact answer with the normal curve, with and without the continuity correction. For n up to 2,000 the answer is computed exactly with whole-number arithmetic; larger n (up to 10 million) uses a high-accuracy floating-point method.
The binomial formula
A binomial setting has a fixed number of trials n, two outcomes per trial (success or failure), independent trials, and the same success probability p every time. The number of successes X then follows the binomial distribution:
P(X = k) = C(n, k) · pk · (1 − p)n − k
- C(n, k) = n! / (k!(n − k)!) is the number of ways to choose which k of the n trials are successes. For example, C(10, 3) = 120.
- pk is the chance that those k trials are all successes.
- (1 − p)n − k is the chance that the other n − k trials are all failures.
The mean is μ = np, the variance is σ² = np(1 − p) and the standard deviation is the square root of the variance. With n = 10 and p = 0.3: μ = 3, σ² = 2.1 and σ ≈ 1.449.
Exactly, at most, at least: which probability to use
| Words in the problem | Symbol | How to get it |
|---|---|---|
| exactly k | P(X = k) | the formula once |
| at most k, no more than k | P(X ≤ k) | add P(X = 0) to P(X = k) |
| fewer than k, less than k | P(X < k) | P(X ≤ k − 1) |
| at least k, k or more | P(X ≥ k) | 1 − P(X ≤ k − 1) |
| more than k | P(X > k) | 1 − P(X ≤ k) |
| between a and b (inclusive) | P(a ≤ X ≤ b) | P(X ≤ b) − P(X ≤ a − 1) |
For a discrete count, “fewer than 3” and “at most 3” are different questions: one stops at 2, the other includes 3. The complement rule saves work whenever the other side has fewer terms; “at least 1” is always 1 − P(X = 0).
Worked examples
Example 1: exactly 3 successes
A basketball player makes 30% of her three-point shots. What is the chance she makes exactly 3 of 10? C(10, 3) = 120, 0.33 = 0.027 and 0.77 = 0.0823543. Multiply: 120 × 0.027 × 0.0823543 = 0.266828, or about 0.2668.
Example 2: at least 3 sixes in 20 rolls
Here n = 20 and p = 1/6. “At least 3” has 18 terms, so use the complement: P(X = 0) = 0.0261, P(X = 1) = 0.1043 and P(X = 2) = 0.1982 add to P(X ≤ 2) = 0.3287. So P(X ≥ 3) = 1 − 0.3287 = 0.6713. The mean is 20 × 1/6 ≈ 3.33 sixes.
Example 3: at most 2 defective parts
A machine makes 8% defective parts and you inspect 50. P(X ≤ 2) = 0.0155 + 0.0672 + 0.1433 = 0.2260. On average you would find np = 4 defects, so finding 2 or fewer happens a little less than a quarter of the time.
Example 4: between 45 and 55 heads, with the normal approximation
Flip a fair coin 100 times. The exact P(45 ≤ X ≤ 55) is 0.7287. With μ = 50 and σ = 5, the normal approximation with the continuity correction uses 44.5 to 55.5: Φ(1.1) − Φ(−1.1) = 0.7287, almost the same. Without the correction (45 to 55, z = ±1) it gives 0.6827, which is far off.
Normal approximation and the continuity correction
When np and n(1 − p) are both at least 10 (some textbooks use 5), the binomial distribution is close to a normal curve with mean np and standard deviation √(np(1 − p)). Because X takes only whole numbers, widen each value by half a unit: P(X ≤ k) becomes P(Y < k + 0.5), P(X ≥ k) becomes P(Y > k − 0.5), and P(X = k) becomes P(k − 0.5 < Y < k + 0.5). The calculator shows the size of the error for your numbers, so you can see when the shortcut is safe.
Common mistakes
| Mistake | Fix |
|---|---|
| Using the formula once for “at least” | Add every term in the range, or use 1 minus the complement. |
| Mixing up < and ≤ | “Fewer than 4” is X ≤ 3; “at least 4” includes 4. |
| Forgetting C(n, k) | pk(1 − p)n − k is the chance of one particular order; multiply by the number of orders. |
| Using the binomial when trials are not independent | Drawing without replacement from a small group changes p each time; that is a hypergeometric problem. |
| Skipping the continuity correction | Add or subtract 0.5 before finding z, as in Example 4. |
Checking homework with the calculator
Do the problem by hand first, then enter the same n, p and k and choose the matching tab. If your answer differs, open the steps and the distribution table: a wrong single term points to C(n, k) or a power, while a wrong sum usually means the range started or stopped one value off.
Binomial probability practice quiz: 10 questions
Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.
C(4, 2) = 6 orders, each with probability 0.5⁴ = 0.0625, so 6 × 0.0625 = 0.375.
C(8, 3) = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 56. The value 336 forgets to divide by 3!.
Variance = np(1 − p) = 40 × 0.25 × 0.75 = 7.5. Its square root, about 2.74, is the standard deviation.
X ≤ 20 includes the bar for 20, which runs from 19.5 to 20.5, so use P(Y < 20.5).
Without replacement, the chance of an ace changes after each draw, so the trials are not independent with a constant p. That count is hypergeometric.
Quiz complete
Frequently asked questions
What is binomial probability?
It is the probability of getting a certain number of successes in a fixed number of independent yes-or-no trials that all have the same success probability, such as the number of heads in 10 coin flips.
How do you calculate binomial probability without a calculator?
Work out C(n, k) with the factorial formula or Pascal’s triangle, raise p to the power k and 1 − p to the power n − k, and multiply the three numbers. For 6 heads in 10 fair flips: 210 × 0.510 = 0.2051.
How do you find an “at least” binomial probability?
Use the complement: P(X ≥ k) = 1 − P(X ≤ k − 1). For at least one success, P(X ≥ 1) = 1 − (1 − p)n; with p = 0.3 and n = 10 that is 1 − 0.710 = 0.9718.
How do you find a binomial probability between two numbers?
Add the single probabilities from the lower value to the upper value, or subtract cumulative values: P(a ≤ X ≤ b) = P(X ≤ b) − P(X ≤ a − 1). Check whether the problem includes the endpoints.
How do you read a binomial probability table?
Find the block for your n, the column for your p and the row for your k. Most printed tables give P(X = k) or P(X ≤ k), so read the heading. The distribution table in the calculator lists both, plus P(X ≥ k), for any n and p, not only the values printed in a textbook.
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