Binomial Distribution: Formula, Mean, Variance and Examples

Counting Successes

A student guesses all the answers in a five-question multiple-choice test. What is the chance of getting at least three right? A factory knows that 5% of its bulbs are faulty. If a customer buys 20, how likely is it that none are faulty? Questions like these, about the number of "successes" in a fixed number of independent trials, are answered by the binomial distribution, one of the most important probability models in statistics. This tutorial covers it for Further Mathematics, STA 111 and MTH 102: the conditions for a binomial experiment, the probability formula, cumulative probabilities, the mean, variance and standard deviation, and worked examples from coins, dice, tests and quality control. It builds on the basic probability tutorial and on combinations.

When Is a Situation Binomial?

A random variable X follows a binomial distribution if all four of these conditions hold:

There is a fixed number of trials, n.
Each trial has only two outcomes, called "success" and "failure".
The probability of success, p, is the same for every trial.
The trials are independent: the result of one does not affect another.

Then X, the number of successes, is written X ~ B(n, p). The probability of failure is q = 1 − p.

Examples and non-examples

Binomial: the number of heads in 10 tosses of a coin, the number of sixes in 12 rolls of a die, or the number of faulty items in a large batch sampled at random.
Not binomial: the number of tosses until the first head (the number of trials is not fixed), or drawing cards from a pack without replacement (the probability changes after each draw, so the trials are not independent).

The Binomial Probability Formula

The probability of exactly r successes in n trials is:

P(X = r) = ⁿCᵣ × pʳ × qⁿ⁻ʳ, for r = 0, 1, 2, …, n

Here's why the formula has this shape:

pʳ qⁿ⁻ʳ is the probability of any one particular sequence with r successes and n − r failures, such as SSFFF.
ⁿCᵣ = n! ÷ (r!(n − r)!) counts how many such sequences there are, that is, how many ways the r successes can be placed among the n trials.

The name comes from the binomial theorem: the probabilities are exactly the terms in the expansion of (q + p)ⁿ, which add up to 1ⁿ = 1.

Worked Examples

Worked example 1: tossing a coin

A fair coin is tossed 6 times. Find the probability of exactly 4 heads.

X ~ B(6, ½). P(X = 4) = ⁶C₄ × (½)⁴ × (½)² = 15 × 1/64 = 15/64 ≈ 0.234.

Worked example 2: a general calculation

If X ~ B(10, 0.3), find P(X = 3).

P(X = 3) = ¹⁰C₃ × 0.3³ × 0.7⁷ = 120 × 0.027 × 0.0823543 ≈ 0.267.

Worked example 3: "at least one"

A fair die is rolled 4 times. Find the probability of at least one six.

"At least one" is best found using the complement, "none":

P(X ≥ 1) = 1 − P(X = 0) = 1 − (5/6)⁴ = 1 − 625/1296 ≈ 0.518.

So it is slightly more likely than not to get at least one six in four rolls.

Worked example 4: guessing in a test

A test has 5 multiple-choice questions, each with 4 options. A student guesses every answer. Find the probability that the student gets at least 3 correct.

X ~ B(5, 0.25). P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5):

P(X = 3) = ⁵C₃ × 0.25³ × 0.75² = 10 × 0.015625 × 0.5625 ≈ 0.0879
P(X = 4) = ⁵C₄ × 0.25⁴ × 0.75 = 5 × 0.00390625 × 0.75 ≈ 0.0146
P(X = 5) = 0.25⁵ ≈ 0.0010

P(X ≥ 3) ≈ 0.104, about a 10% chance. Guessing is not a good strategy.

Worked example 5: quality control

5% of the bulbs made by a factory are faulty. A random sample of 20 is taken. Find the probability that (a) none is faulty and (b) at most one is faulty.

X ~ B(20, 0.05).

(a) P(X = 0) = 0.95²⁰ ≈ 0.358.

(b) P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.358 + 20 × 0.05 × 0.95¹⁹ ≈ 0.358 + 0.377 = 0.736.

Cumulative Probabilities and Key Phrases

Many questions ask for a range of values. Translate the words carefully:

"At most 2" means P(X ≤ 2) = P(0) + P(1) + P(2).
"Fewer than 2" means P(X ≤ 1).
"At least 2" means P(X ≥ 2) = 1 − P(X ≤ 1).
"More than 2" means P(X ≥ 3) = 1 − P(X ≤ 2).

Using the complement often saves a lot of work. For example, with n = 10, "at least 1" needs ten terms directly, but only one term with the complement.

Worked example 6

If X ~ B(10, 0.3), find P(X ≤ 1) and P(X ≥ 2).

P(X = 0) = 0.7¹⁰ ≈ 0.0282, and P(X = 1) = 10 × 0.3 × 0.7⁹ ≈ 0.1211.

P(X ≤ 1) ≈ 0.149, so P(X ≥ 2) ≈ 1 − 0.149 = 0.851.

Mean, Variance and Standard Deviation

For X ~ B(n, p):

Mean (expected value): E(X) = np
Variance: Var(X) = npq = np(1 − p)
Standard deviation: σ = √(npq)

The mean is intuitive: if 30% of trials succeed on average, then 10 trials give 3 successes on average.

Worked example 7

For X ~ B(10, 0.3), find the mean, variance and standard deviation.

Mean = 10 × 0.3 = 3. Variance = 10 × 0.3 × 0.7 = 2.1. Standard deviation = √2.1 ≈ 1.45.

Worked example 8: working backwards

A binomial distribution has mean 6 and variance 4.2. Find n and p.

np = 6 and npq = 4.2, so q = 4.2 ÷ 6 = 0.7. Then p = 0.3, and n = 6 ÷ 0.3 = 20.

The Shape of the Distribution

When p = 0.5, the distribution is symmetric about its mean.
When p is small, it is skewed to the right, with most of the probability on small values.
When p is large, it is skewed to the left.
As n increases, the distribution becomes more bell-shaped. For large n, it can be approximated by the normal distribution, and when n is large and p is small, by the Poisson distribution with mean np.

Common Mistakes with the Binomial Distribution

Forgetting the ⁿCᵣ factor, which counts only one arrangement of the successes.
Misreading "at least" and "at most", or "fewer than" and "more than". Write the inequality before calculating.
Using the binomial model without replacement from a small population, where the trials are not independent.
Calculating every term when the complement is much quicker.
Using np for the standard deviation. The variance is npq, and the standard deviation is its square root.

Binomial Distribution: Practice Questions

A fair coin is tossed 5 times. Find the probability of exactly 3 heads.
If X ~ B(8, 0.2), find P(X = 0) and P(X = 2).
A die is rolled 3 times. Find the probability of exactly one six.
60% of the students in a school pass mathematics. Five students are chosen at random. Find the probability that all five passed.
Find the mean and variance of X ~ B(25, 0.4).
A binomial distribution has mean 4 and variance 3. Find n and p.

Answers: 1. ⁵C₃ × (½)⁵ = 10/32 = 5/16 ≈ 0.313. 2. P(X = 0) = 0.8⁸ ≈ 0.168, and P(X = 2) = 28 × 0.04 × 0.8⁶ ≈ 0.294. 3. 3 × (1/6) × (5/6)² = 75/216 ≈ 0.347. 4. 0.6⁵ ≈ 0.078. 5. Mean = 10 and variance = 25 × 0.4 × 0.6 = 6. 6. q = 3/4, so p = 1/4 and n = 16.

Conclusion

The binomial distribution models the number of successes in n independent trials, each with the same probability p of success: X ~ B(n, p), with P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ. Translate phrases such as "at least" and "at most" into inequalities, and use the complement to save work. The mean is np, the variance npq and the standard deviation √(npq). The model applies to coin tosses, dice, guessing in tests and quality control, and for large n it leads naturally to the normal distribution.