Maths
What is the binomial distribution, and how do you use the formula?
Quick answer
The binomial distribution B(n, p) gives the probability of k successes in n independent trials, each with success probability p: P(X = k) = ⁿCₖ pᵏ (1 − p)ⁿ⁻ᵏ. Its mean is np.
In one sentence
The binomial distribution tells you how likely it is to get exactly 0, 1, 2, … successes when you repeat the same yes-or-no experiment a fixed number of times.
Explained
A footballer scores 80% of her penalties. If she takes 5 penalties, how likely is she to score exactly 4? Each penalty is a trial with two outcomes, success (goal) or failure (miss). The number of goals, X, can be 0, 1, 2, 3, 4 or 5, and the binomial distribution gives the probability of each value.
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You can use it when four conditions hold. A memory aid is BINS:
- Binary: each trial has only two outcomes, success or failure.
- Independent: one trial does not affect another.
- Number of trials is fixed: you know n in advance.
- Same probability: the chance of success p is the same in every trial.
Then we write X ~ B(n, p): "X follows a binomial distribution with n trials and success probability p". The formula is
P(X = k) = ⁿCₖ × pᵏ × (1 − p)ⁿ⁻ᵏ
Each part has a job. pᵏ is the chance of k successes, (1 − p)ⁿ⁻ᵏ is the chance of the n − k failures, and ⁿCₖ ("n choose k") counts how many different orders those successes can come in. For 5 penalties with exactly 4 goals, the miss could be the 1st, 2nd, 3rd, 4th or 5th penalty, so ⁵C₄ = 5. You can find ⁿCₖ with the nCr button on a calculator, from Pascal's triangle, or with n! ÷ (k!(n − k)!).
So P(X = 4) = 5 × 0.8⁴ × 0.2¹ = 5 × 0.4096 × 0.2 = 0.4096.
Three more facts are used all the time:
- Mean (expected value): μ = np. Out of 5 penalties she expects 5 × 0.8 = 4 goals on average.
- Variance: σ² = np(1 − p), and the standard deviation is σ = √(np(1 − p)). See standard deviation.
- Cumulative probability: P(X ≤ k) adds up P(X = 0) to P(X = k). Tables and calculators give these directly.
The binomial distribution is discrete: X can only be a whole number from 0 to n. It is symmetric when p = 0.5, skewed to the right when p is small and skewed to the left when p is close to 1. When n is large and p is not too close to 0 or 1, its shape gets close to the bell curve of the normal distribution.
Examples
The whole distribution for 4 trials, with two different values of p:
| k | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| B(4, 0.5): P(X = k) | 0.0625 | 0.25 | 0.375 | 0.25 | 0.0625 |
| B(4, 0.2): P(X = k) | 0.4096 | 0.4096 | 0.1536 | 0.0256 | 0.0016 |
Each row adds up to 1. The first row is symmetric; the second is skewed, with most of the probability at small k.
Worked example: roll a fair dice 6 times. What is the probability of exactly 2 sixes? Here n = 6, p = 1/6 and k = 2, with ⁶C₂ = 15.
P(X = 2) = 15 × (1/6)² × (5/6)⁴ = 15 × 625 ÷ 46,656 = 9,375 ÷ 46,656 ≈ 0.2009
Worked example: "at least one". X ~ B(5, 0.2). P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.8⁵ = 1 − 0.32768 = 0.67232.
Common mistakes
- Forgetting ⁿCₖ. 0.8⁴ × 0.2 is the probability of one particular order only. Fix: multiply by the number of orders.
- Mixing up "less than" and "at most". P(X < 3) = P(X ≤ 2), but P(X ≤ 3) includes 3. P(X ≥ 3) = 1 − P(X ≤ 2). Fix: write out the values of k you need.
- Using it when p changes. Drawing cards without putting them back changes p each time, so it is not binomial (it is hypergeometric). Fix: check BINS first.
- Mixing up the mean and the variance. Fix: the mean is np; the variance is np(1 − p), which is never bigger than the mean.
Step by step
Step by step
- Check the conditionsTwo outcomes, independent trials, fixed n, constant p.
- Name n, p and kWrite X ~ B(n, p) and the value or values of k you need.
- Choose the right probabilityExactly k: use the formula. At most k: add up or use a cumulative table. At least k: 1 − P(X ≤ k − 1).
- CalculateP(X = k) = ⁿCₖ pᵏ (1 − p)ⁿ⁻ᵏ, or use your calculator's binomial functions.
- Check it makes senseIs the answer between 0 and 1? Is it biggest near the mean np?
Practice questions
The binomial distribution
Try each one, then check your answer.
- Is "the number of sixes in 20 rolls of a fair dice" binomial? If so, give n and p.
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Yes: two outcomes, independent rolls, fixed n, constant p. n = 20, p = 1/6 - Is "the number of cards you draw until you get an ace" binomial?
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No: the number of trials is not fixed. - X ~ B(4, 0.5). Find P(X = 2).
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⁴C₂ = 6, so 6 × 0.5² × 0.5² = 6 × 0.0625 = 0.375 - X ~ B(5, 0.2). Find P(X = 0).
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0.8⁵ = 0.32768 - X ~ B(5, 0.2). Find P(X ≥ 1).
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1 − 0.32768 = 0.67232 - X ~ B(6, 0.3). Find P(X = 2).
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15 × 0.3² × 0.7⁴ = 15 × 0.09 × 0.2401 ≈ 0.3241 - X ~ B(10, 0.5). Find P(X = 10).
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0.5¹⁰ = 1/1024 ≈ 0.000977 - X ~ B(3, 0.4). Find P(X ≤ 1).
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P(0) = 0.6³ = 0.216; P(1) = 3 × 0.4 × 0.6² = 0.432. Total 0.648 - X ~ B(20, 0.25). Find the mean.
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np = 20 × 0.25 = 5 - X ~ B(20, 0.25). Find the variance and the standard deviation.
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Variance = 20 × 0.25 × 0.75 = 3.75; SD = √3.75 ≈ 1.936 - X ~ B(8, 0.5). Find P(X = 4).
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⁸C₄ = 70, so 70 × 0.5⁸ = 70 ÷ 256 ≈ 0.2734 - X ~ B(4, 0.1). Find P(X ≥ 2).
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P(0) = 0.6561, P(1) = 4 × 0.1 × 0.729 = 0.2916. 1 − 0.6561 − 0.2916 = 0.0523 - X ~ B(n, 0.4) has mean 12. Find n.
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0.4n = 12, so n = 30 - X ~ B(n, p) has mean 6 and variance 4.2. Find n and p.
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1 − p = 4.2 ÷ 6 = 0.7, so p = 0.3 and n = 6 ÷ 0.3: n = 20, p = 0.3 - What is the most likely value of X ~ B(10, 0.3)?
Show answer
P(2) ≈ 0.2335, P(3) ≈ 0.2668, P(4) ≈ 0.2001, so the most likely value is 3
Word problems
The binomial distribution in real life
Try each one, then check your answer.
- A footballer scores 80% of her penalties. She takes 5. What is the probability that she scores all 5?
Show answer
0.8⁵ = 0.32768 - Seeds in a packet each have a 90% chance of growing. You plant 10. How many do you expect to grow, and what is the probability that all 10 grow?
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Mean 10 × 0.9 = 9 seeds; P(all 10) = 0.9¹⁰ ≈ 0.3487 - A quiz has 10 questions with 4 options each. Ana guesses every answer. How many does she expect to get right, and what is the probability that she gets none right?
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Mean 10 × 0.25 = 2.5; P(0) = 0.75¹⁰ ≈ 0.0563 - 2% of light bulbs from a factory are faulty. A box holds 50. What is the probability that the box has at least one faulty bulb?
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1 − 0.98⁵⁰ ≈ 1 − 0.3642 = 0.6358 - A family has 3 children, and each child is equally likely to be a girl or a boy. What is the probability of exactly 2 girls?
Show answer
³C₂ × 0.5³ = 3 × 0.125 = 0.375
Questions people ask
What is the binomial distribution used for?
For counting successes in repeated yes-or-no trials: goals from penalties, faulty items in a batch, correct guesses in a test, patients who respond to a treatment. It is also the basis of many simple hypothesis tests.
Is the binomial distribution discrete or continuous?
Discrete. X counts successes, so it can only be 0, 1, 2, … up to n. The normal distribution, by contrast, is continuous.
How do you use a binomial distribution table?
Most tables give cumulative probabilities P(X ≤ k) for chosen n and p. Find your n and p, then read the row for k. For P(X = k), subtract: P(X ≤ k) − P(X ≤ k − 1). For P(X ≥ k), use 1 − P(X ≤ k − 1).
How do you do the binomial distribution on a TI-84?
Press 2nd, VARS (DISTR). binompdf(n, p, k) gives P(X = k), and binomcdf(n, p, k) gives P(X ≤ k). For X ~ B(5, 0.8), binompdf(5, 0.8, 4) returns 0.4096.
How do you find the binomial distribution in Excel?
=BINOM.DIST(k, n, p, FALSE) gives P(X = k), and =BINOM.DIST(k, n, p, TRUE) gives P(X ≤ k). For example, =BINOM.DIST(4, 5, 0.8, FALSE) gives 0.4096.
What is the negative binomial distribution?
It turns the question around: instead of fixing the number of trials and counting successes, it counts how many trials (or failures) you need before reaching a fixed number of successes. It is used, for example, for "how many calls until the third sale?".
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The binomial distribution
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