Case 1
- Number of trials (n)
- 10
- Number of successes (k)
- 5
- Success probability per trial (p)
- 0.5
0.246094
Open with these valuesBinomial Probability Calculator
0.246094
Result: 0.246094This returns the chance of exactly k successes in n trials — one term of the distribution, not a running total. Five heads in ten fair flips is 0.246094, so the single most likely count still happens less than a quarter of the time.
The numbers at a glance
Held fixed: Number of trials (n) 10, Number of successes (k) 5.
| Success probability per trial (p) | Result |
|---|---|
| 0.300000 | 0.102919 |
| 0.400000 | 0.200658 |
| 0.500000Your value | 0.246094 |
| 0.600000 | 0.200658 |
| 0.700000 | 0.102919 |
Worked examples
Case 2
- Number of trials (n)
- 20
- Number of successes (k)
- 6
- Success probability per trial (p)
- 0.3
0.191639
Open with these valuesCase 3
- Number of trials (n)
- 12
- Number of successes (k)
- 4
- Success probability per trial (p)
- 0.25
0.193578
Open with these valuesHow it's calculated
P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ
- StepEnter how many trials you run — the number must be fixed in advance.
- StepEnter the exact number of successes you are asking about.
- StepEnter the chance of success on a single trial, as a decimal.
- ResultRead the probability of exactly that count.
What this number means
The binomial distribution answers a question about a count, not about a sequence: across n independent trials that each succeed with probability p, how likely is exactly k successes? The formula is two ideas multiplied together. Hold one particular pattern fixed — say heads on the first five of ten flips and tails on the rest. Its probability is p taken k times and (1 − p) taken n − k times, which at the values set here is 0.5¹⁰ = 0.0009765625. Every other pattern with five heads carries exactly the same probability, because it holds the same factors in a different order. All that remains is to count the patterns, and that is what the binomial coefficient does: C(10, 5) = 252 ways of choosing which five flips come up heads. Multiplied out, 252 × 0.0009765625 = 0.246094. That structure explains why the conditions are not small print. A constant p is what gives every pattern the same probability, and independence is what allows the factors to be multiplied at all; without either, the coefficient counts patterns that are no longer interchangeable and the product means nothing. What the result never says is which trials succeeded, or that k successes did occur — only how much probability sits on that one count.
A binomial coefficient times two powers
C(n, k) counts which of the trials succeed, and pᵏ × (1 − p)ⁿ⁻ᵏ weighs one such pattern. Five heads in ten fair flips: 252 × 0.5¹⁰ = 0.246094.
Constant p and independent trials are required
The formula needs a fixed number of trials, two outcomes each, the same success probability every time, and no trial influencing another. Drawing without replacement breaks the last two, and the hypergeometric distribution fits that case instead.
Exactly k, never k or more
The result is a single term of the distribution, P(X = k). For at least 8 heads in 10 flips you add P(8) + P(9) + P(10), each from the same formula.
Commonly misread
P(X = 5) at n = 10 and p = 0.5 is the chance of five heads or fewer.
Five or fewer comes to 0.623047; exactly five is 0.246094. This calculator always returns the single term.
The most likely count is the one that usually happens.
Five heads in ten fair flips is the peak of the distribution and still only 0.246094, under a quarter. A most likely value is not a likely value.
With p = 0.9, five successes out of five should be close to 0.9.
Every one of the five trials has to succeed, so the chance is 0.9 taken five times over: 0.590490. Repeated multiplication pulls the number down fast.
Reference table
| n, k, p | Case | P(X = k) |
|---|---|---|
| 10, 5, 0.5 | five heads in ten fair flips | 0.246094 |
| 10, 0, 0.5 | no heads at all | 0.000977 |
| 12, 4, 0.25 | four of twelve at 25 % | 0.193578 |
| 20, 6, 0.3 | six of twenty at 30 % | 0.191639 |
| 5, 5, 0.9 | five for five at 90 % | 0.590490 |
Questions
How do I calculate a binomial probability?
Multiply the binomial coefficient by the success and failure probabilities: P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ. Exactly five heads in ten fair coin flips is 252 × 0.5¹⁰ = 0.246094, about 24.6 %.
Does this give exactly k or at least k successes?
Exactly k — a single term of the distribution. For at least k or at most k you add the individual probabilities: the chance of at least 8 heads in 10 flips is P(8) + P(9) + P(10), each from the same formula.
When may I use the binomial formula?
Four conditions must hold: a fixed number of trials, only two outcomes per trial, the same success probability on every trial, and independent trials. If you draw without replacement from a small population, the hypergeometric distribution fits better.
What is the binomial distribution?
It describes the number of successes in a fixed number of independent trials that share one success probability. It applies whenever you repeat a yes-or-no experiment: coin flips, free throws, defective items coming off a line.
Why is the result always between 0 and 1?
A probability is a share of all possible outcomes, so it can never fall below 0 or rise above 1. Multiply by 100 to read it as a percentage. Summed over every k from 0 to n the terms come to exactly 1, because one of those counts has to occur.
Sources and last check
Information, not professional advice.