Binomial Distribution Calculator (Probability Table)

This binomial distribution calculator finds binomial probabilities: exactly k successes, at most k, fewer than k, at least k or more than k, out of n independent trials with success probability p. It shows the formula with your numbers, all five probabilities at once, the mean, variance and standard deviation, and, for up to 30 trials, a table of every probability and cumulative probability.

It’s free and needs no sign-up.

How to use it

1. Enter the number of trials n.
2. Enter the probability of success on each trial p, as a decimal (0.5, not 50%).
3. Next, enter the number of successes k and choose which probability you want.
4. Press Calculate.

when to use the binomial distribution calculator

A situation is binomial when all four conditions hold:

1. A fixed number of trials n.
2. Each trial has two outcomes: success or failure.
3. The probability of success p is the same on every trial.
4. The trials are independent.

Coin flips, multiple-choice guessing, defective items in a sample and free throws all fit.

Formula

The probability of exactly k successes is

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

where ( binom{n}{k} ) counts the ways to choose which k trials are the successes (see the combination calculator). Cumulative probabilities add these up: ( P(X le k) = P(X = 0) + cdots + P(X = k) ).

The mean and standard deviation are

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

Worked example: 3 heads in 10 flips

As loaded above, with n = 10, p = 0.5, k = 3:

$$P(X = 3) = binom{10}{3}(0.5)^3(0.5)^7 = 120 times frac{1}{1024} approx 0.1172$$

The calculator also gives ( P(X le 3) = 0.1719 ) and ( P(X ge 3) = 0.9453 ). The mean is ( 10 times 0.5 = 5 ) heads.

Worked example: quality control

A machine produces 5% defective parts. In a sample of 20, what is the probability of at least 2 defective ones? Use the complement:

$$P(X ge 2) = 1 - P(X = 0) - P(X = 1) = 1 - 0.95^{20} - 20(0.05)(0.95)^{19}$$

$$= 1 - 0.3585 - 0.3774 = 0.2642$$

So about 26% of samples will contain two or more defective parts. Enter n = 20, p = 0.05, k = 2 and choose “P(X ≥ k)”.

Worked example: guessing on a test

A quiz has 5 questions with 5 choices each. Guessing every answer, the chance of getting none right is ( 0.8^5 = 0.32768 ), about 33%. As a result, the expected score is ( 5 times 0.2 = 1 ) question.

“At least”, “at most” and the others

Words	Probability
exactly k	( P(X = k) )
at most k, no more than k	( P(X le k) )
fewer than k	( P(X < k) = P(X le k – 1) )
at least k, k or more	( P(X ge k) = 1 – P(X le k – 1) )
more than k	( P(X > k) = 1 – P(X le k) )

Normal approximation

When ( np ge 10 ) and ( n(1 – p) ge 10 ), the binomial distribution is close to a normal distribution with the same mean and standard deviation, so you can use a z-score (with a continuity correction of ±0.5). This calculator computes the exact binomial probabilities, so you don’t need the approximation, even for large n.

Common mistakes

• Mixing up “at least” and “at most”.
• Using percentages for p. Enter 0.05, not 5.
• Using the binomial when trials aren’t independent, such as drawing cards without replacement from a small deck.

Step-by-step

1. Check the four conditions: fixed n, two outcomes, constant p, independent trials.
2. Identify n, p and the values of k you need.
3. Translate the words (“at least”, “at most”) into an inequality.
4. Use the formula for each k, or the complement if it is shorter, and add.

Worked example: free throws

A basketball player makes 80% of free throws. In 10 attempts, the expected number made is ( mu = 10 times 0.8 = 8 ), with standard deviation ( sqrt{10 times 0.8 times 0.2} = sqrt{1.6} approx 1.26 ). Also, the chance of making all 10 is ( 0.8^{10} approx 0.107 ), about 11%.

Worked example: a fair coin, 12 tosses

What is the chance of 10 or more heads? With ( p = frac12 ) every sequence has probability ( frac{1}{4096} ), so

$$P(X ge 10) = frac{binom{12}{10} + binom{12}{11} + binom{12}{12}}{4096} = frac{66 + 12 + 1}{4096} = frac{79}{4096} approx 0.0193$$

Under 2%, so 10 heads in 12 tosses would make you suspect the coin.

The shape of the distribution

When ( p = 0.5 ) the distribution is symmetric about ( frac{n}{2} ). When p is small it is skewed right (most values near 0, a long tail to the right); when p is close to 1 it is skewed left. As n grows, it looks more and more like a bell curve, which is why the normal approximation works for large n.

Mean and variance: why np?

The number of successes is a sum of n independent 0/1 variables, each with mean p and variance ( p(1 – p) ). Means and variances of independent variables add, so the total has mean np and variance ( np(1 – p) ). See the standard deviation formula for how variance and standard deviation relate.

Practice problems

1. ( n = 8 ), ( p = 0.25 ): find ( P(X = 2) ).
2. ( n = 12 ), ( p = 0.5 ): find ( P(X ge 10) ).
3. ( n = 5 ), ( p = 0.2 ): find ( P(X = 0) ).

Answers: 1. ( 28 times 0.25^2 times 0.75^6 approx 0.3115 ). 2. ( frac{79}{4096} approx 0.0193 ). 3. ( 0.8^5 = 0.32768 ).

Probabilities over a range

To find ( P(a le X le b) ), subtract cumulative probabilities: ( P(a le X le b) = P(X le b) – P(X le a – 1) ). For 10 fair coin flips, ( P(3 le X le 5) = P(X le 5) – P(X le 2) = 0.6230 – 0.0547 = 0.5684 ). The calculator’s table lists every ( P(X le k) ) for up to 30 trials, so you can read both values straight off it.

Related guides

• Permutation and Combination Calculator
• Z-Score
• Standard Deviation Formula
• Mean, Median and Mode

Further reading

• Binomial distribution (Wikipedia) — formula, properties and approximations.

FAQ

What is the binomial probability formula?

( P(X = k) = binom{n}{k} p^k (1 – p)^{n-k} ): the number of ways to get k successes times the probability of any one such sequence.

What is the mean of a binomial distribution?

( mu = np ), the expected number of successes.

How do you find “at least” binomial probabilities?

Use the complement: ( P(X ge k) = 1 – P(X le k – 1) ).

What is the difference between binomial pdf and cdf?

The pdf (or pmf) gives ( P(X = k) ); the cdf gives the cumulative ( P(X le k) ). The calculator gives both.

What is the difference between binomial and normal distributions?

The binomial counts successes, so it takes whole-number values from 0 to n. In contrast, the normal is continuous. For large n, the binomial is approximately normal.

What is the most likely number of successes?

The mode is usually near np: it is the whole number ( lfloor (n + 1)p rfloor ) (and sometimes the one just below it too).