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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.

Ages 15–18Secondary

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.

  • football
  • flower
  • lamp
Penalties scored, seeds that grow and faulty light bulbs can all follow a binomial distribution.

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

  1. Check the conditionsTwo outcomes, independent trials, fixed n, constant p.
  2. Name n, p and kWrite X ~ B(n, p) and the value or values of k you need.
  3. 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).
  4. CalculateP(X = k) = ⁿCₖ pᵏ (1 − p)ⁿ⁻ᵏ, or use your calculator's binomial functions.
  5. 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.

  1. Is "the number of sixes in 20 rolls of a fair dice" binomial? If so, give n and p.
    Show answer
    Yes: two outcomes, independent rolls, fixed n, constant p. n = 20, p = 1/6
  2. Is "the number of cards you draw until you get an ace" binomial?
    Show answer
    No: the number of trials is not fixed.
  3. X ~ B(4, 0.5). Find P(X = 2).
    Show answer
    ⁴C₂ = 6, so 6 × 0.5² × 0.5² = 6 × 0.0625 = 0.375
  4. X ~ B(5, 0.2). Find P(X = 0).
    Show answer
    0.8⁵ = 0.32768
  5. X ~ B(5, 0.2). Find P(X ≥ 1).
    Show answer
    1 − 0.32768 = 0.67232
  6. X ~ B(6, 0.3). Find P(X = 2).
    Show answer
    15 × 0.3² × 0.7⁴ = 15 × 0.09 × 0.2401 ≈ 0.3241
  7. X ~ B(10, 0.5). Find P(X = 10).
    Show answer
    0.5¹⁰ = 1/1024 ≈ 0.000977
  8. X ~ B(3, 0.4). Find P(X ≤ 1).
    Show answer
    P(0) = 0.6³ = 0.216; P(1) = 3 × 0.4 × 0.6² = 0.432. Total 0.648
  9. X ~ B(20, 0.25). Find the mean.
    Show answer
    np = 20 × 0.25 = 5
  10. X ~ B(20, 0.25). Find the variance and the standard deviation.
    Show answer
    Variance = 20 × 0.25 × 0.75 = 3.75; SD = √3.75 ≈ 1.936
  11. X ~ B(8, 0.5). Find P(X = 4).
    Show answer
    ⁸C₄ = 70, so 70 × 0.5⁸ = 70 ÷ 256 ≈ 0.2734
  12. X ~ B(4, 0.1). Find P(X ≥ 2).
    Show answer
    P(0) = 0.6561, P(1) = 4 × 0.1 × 0.729 = 0.2916. 1 − 0.6561 − 0.2916 = 0.0523
  13. X ~ B(n, 0.4) has mean 12. Find n.
    Show answer
    0.4n = 12, so n = 30
  14. X ~ B(n, p) has mean 6 and variance 4.2. Find n and p.
    Show answer
    1 − p = 4.2 ÷ 6 = 0.7, so p = 0.3 and n = 6 ÷ 0.3: n = 20, p = 0.3
  15. 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.

  1. 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
  2. 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?
    Show answer
    Mean 10 × 0.9 = 9 seeds; P(all 10) = 0.9¹⁰ ≈ 0.3487
  3. 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?
    Show answer
    Mean 10 × 0.25 = 2.5; P(0) = 0.75¹⁰ ≈ 0.0563
  4. 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?
    Show answer
    1 − 0.98⁵⁰ ≈ 1 − 0.3642 = 0.6358
  5. 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?".

Quick quiz

The binomial distribution

Check what you know with 5 quick questions.

Question 1 of 5

Which of these is NOT a condition for a binomial distribution?

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