Binomial Distribution

A binomial distribution gives the probability of getting exactly k successes in n independent trials, when each trial has the same probability of success p. The binomial formula is P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}. Its mean is np, and its standard deviation is \sqrt{np(1 - p)}.

Before this lesson: Independent Events, Permutations and Combinations and Expected Value

When is a random variable binomial?

A binomial random variable X counts the successes in a set of trials. Check four conditions, often remembered as BINS:

• Binary: each trial has two outcomes, called success and failure.
• Independent: the result of one trial doesn’t change the others.
• Number: the number of trials, n, is fixed in advance.
• Same probability: the chance of success, p, is the same on every trial.

“Success” just means the outcome you are counting. It can be a defective part or a wrong answer.

When you sample without replacement, the trials are not quite independent. You can still treat them as independent if the sample is less than 10% of the population. This is called the 10% condition.

Binomial distribution formula

For a binomial random variable with n trials and success probability p:

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

Here \binom{n}{k}, read “n choose k,” counts the ways to place k successes among n trials. Each of those orders has probability p^k (1 - p)^{n - k}. The mean and standard deviation are:

\mu_X = np \qquad \sigma_X = \sqrt{np(1 - p)}

Calculator check: on a TI-84, press 2nd, VARS (DISTR). Use binompdf(n, p, k) for P(X = k) and binomcdf(n, p, k) for P(X \le k). In Desmos, type binomialdist(n, p), then use .pdf(k) or .cdf(k) on it. Work the formula by hand first, then check.

Binomial distribution examples

Example 1: Is it binomial?

Decide whether X is binomial. (a) Roll a die 10 times; X = the number of 6s. (b) Draw cards from a deck without replacement until you get an ace; X = the number of draws. (c) Pick 5 students from a class of 20 without replacement; X = the number who are left-handed.

\text{(a) } n = 10, \ p = \tfrac{1}{6}

\text{(b) no fixed } n

\text{(c) } \tfrac{5}{20} = 25\%

Answer: only (a) is binomial. In (b) the number of trials isn’t fixed, and in (c) the picks are not independent.

Example 2: Exactly 3 successes

A basketball player makes 70% of her foul shots. She takes 5 shots, and the shots are independent. Find the probability that she makes exactly 3.

n = 5, \ p = 0.7, \ k = 3

\binom{5}{3} = 10

0.7^3 = 0.343

0.3^2 = 0.09

10(0.343)(0.09) = 0.3087

Answer: P(X = 3) = 0.3087. Check: binompdf(5, 0.7, 3) gives 0.3087.

Example 3: At least one and at most two

A student guesses on 10 multiple-choice questions, each with 4 choices. Let X be the number correct, so n = 10 and p = 0.25. Find P(X \ge 1) and P(X \le 2).

P(X = 0) = 0.75^{10} \approx 0.0563

P(X \ge 1) = 1 - P(X = 0)

1 - 0.0563 = 0.9437

P(X = 1) = \binom{10}{1}(0.25)^1(0.75)^9

\approx 10(0.25)(0.07508) \approx 0.1877

P(X = 2) = \binom{10}{2}(0.25)^2(0.75)^8

\approx 45(0.0625)(0.10011) \approx 0.2816

P(X \le 2) = 0.0563 + 0.1877 + 0.2816

= 0.5256

Answer: P(X \ge 1) \approx 0.9437 and P(X \le 2) \approx 0.5256. Check with binomcdf(10, 0.25, 2).

Example 4: Mean and standard deviation

For the guessing student in Example 3, find the mean and standard deviation of X.

\mu_X = np = 10(0.25)

= 2.5

\sigma_X = \sqrt{np(1 - p)} = \sqrt{10(0.25)(0.75)}

= \sqrt{1.875}

\approx 1.37

Answer: on average a guesser gets 2.5 questions right, give or take about 1.37. The histogram shows the whole distribution.

Example 5: Defective phones (word problem)

A factory finds that 8% of its phones have a defect (example data). An inspector picks 20 phones at random from a large day’s production. Find the probability that no more than 1 is defective, and find the mean and standard deviation of the number defective.

n = 20, \ p = 0.08

P(X = 0) = 0.92^{20} \approx 0.1887

P(X = 1) = \binom{20}{1}(0.08)(0.92)^{19}

\approx 20(0.08)(0.2051)

\approx 0.3282

P(X \le 1) = 0.1887 + 0.3282 = 0.5169

\mu_X = 20(0.08) = 1.6

\sigma_X = \sqrt{20(0.08)(0.92)} = \sqrt{1.472}

\approx 1.21

Answer: P(X \le 1) \approx 0.5169, about a 52% chance. The inspector expects 1.6 defective phones, with a standard deviation of about 1.21.

Common mistakes

Related methods

Where n choose k comes from

The coefficients \binom{n}{k} are the same numbers as in the binomial theorem and Pascal’s triangle. That is where the binomial distribution gets its name: the probabilities are the terms of (p + (1 - p))^n.

Using the mean and standard deviation

The mean and standard deviation formulas come from combining random variables. A binomial X is the sum of n independent trials. Each trial has mean p and variance p(1 - p), so the means and the variances add up n times.

Practice problems

1. A coin is flipped 12 times, and X is the number of tails. Is X binomial? If so, give n and p.
2. A die is rolled until a 6 appears, and X is the number of rolls. Is X binomial?
3. X is binomial with n = 4 and p = 0.5. Find P(X = 2).
4. X is binomial with n = 6 and p = 0.2. Find P(X = 0).
5. Use Problem 4. Find P(X ≥ 1).
6. A player makes 70% of her foul shots. Find the probability that she makes all 5 of 5 shots.
7. X is binomial with n = 50 and p = 0.4. Find the mean and standard deviation.
8. About 15% of people are left-handed. In a random group of 8 people, find the probability that exactly 2 are left-handed.
9. X is binomial with n = 3 and p = 0.6. Find the whole probability distribution.
10. A student guesses on 20 questions with 5 choices each. Find the mean and standard deviation of the number correct, and P(X = 0).

\text{2 outcomes, independent, } n = 12, \ p = 0.5

\text{no fixed } n

\binom{4}{2} = 6

0.5^2 \cdot 0.5^2 = 0.0625

6(0.0625) = 0.375

\binom{6}{0} = 1

0.8^6 = 0.262144

1 - 0.262144 = 0.737856

0.7^5 = 0.16807

\mu = 50(0.4) = 20

\sigma = \sqrt{50(0.4)(0.6)} = \sqrt{12}

\sqrt{12} \approx 3.46

\binom{8}{2} = 28

0.15^2 = 0.0225

0.85^6 \approx 0.37715

28(0.0225)(0.37715) \approx 0.2376

P(X = 0) = 0.4^3 = 0.064

P(X = 1) = 3(0.6)(0.4^2) = 0.288

P(X = 2) = 3(0.6^2)(0.4) = 0.432

P(X = 3) = 0.6^3 = 0.216

0.064 + 0.288 + 0.432 + 0.216 = 1

\mu = 20(0.2) = 4

\sigma = \sqrt{20(0.2)(0.8)} = \sqrt{3.2}

\sqrt{3.2} \approx 1.79

P(X = 0) = 0.8^{20} \approx 0.0115

Binomial distribution FAQ

What are the 4 conditions for a binomial distribution?

Each trial has two outcomes, the trials are independent, the number of trials is fixed, and the probability of success is the same on every trial (BINS).

What is the difference between binompdf and binomcdf?

binompdf gives the probability of exactly k successes. binomcdf gives the probability of k or fewer successes.

How do you find the mean of a binomial distribution?

Multiply the number of trials by the probability of success: \mu = np. The standard deviation is \sqrt{np(1 - p)}.

How do you find “at least” binomial probabilities?

Use the complement. For example, P(X \ge 1) = 1 - P(X = 0), and P(X \ge 3) = 1 - P(X \le 2), which is 1 minus binomcdf(n, p, 2).

Summary

• A binomial random variable counts successes in n independent trials with the same success probability p (BINS).
• P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}.
• Mean np; standard deviation \sqrt{np(1 - p)}.
• Use binompdf for “exactly” and binomcdf for “at most”; use the complement for “at least.”
• When sampling without replacement, check that the sample is under 10% of the population.

Previous lesson: Combining Random Variables

Next lesson: Geometric Distribution

Back to the Statistics Index