The Binomial Distribution: Exactly k Successes in n Trials

Quick answer

A binomial setting repeats the same trial n times, independently, and each trial succeeds with the same probability p. The number of successes X then follows the binomial distribution: P(X = k) = C(n, k) p^k (1 − p)^(n − k). The combination counts the ways to choose which k trials succeed, and each of those outcomes has probability p^k (1 − p)^(n − k). Questions about at least or at most add several of these probabilities, often most quickly through the complement.

What you'll learn

Recognize a binomial setting from its four conditions
Compute the probability of exactly k successes
Explain the formula as a count of arrangements times one probability
Find at least and at most probabilities, using the complement
Describe the shape of a binomial distribution

Counting successes

A basketball player makes $70\%$ of her free throws. She shoots $5$ . What is the probability that she makes exactly $4$ ?

Problems like this share four features, and together they make a binomial setting:

Two outcomes. Each trial is a success or a failure.
Independent trials. One shot does not change the chances of another.
A fixed number of trials, $n$ .
The same probability of success, $p$ , on every trial.

The number of successes, $X$ , then has the binomial distribution:

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

For the free throws, $n = 5$ , $p = 0.7$ and $k = 4$ :

P(X = 4) = \binom{5}{4}(0.7)^4(0.3)^1 = 5 \times 0.2401 \times 0.3 \approx 0.360

Why the formula works

Write S for a make, a success, and F for a miss, a failure. One way to make exactly $4$ is SSSSF. The shots are independent, so its probability is a product: $0.7 \times 0.7 \times 0.7 \times 0.7 \times 0.3 = 0.7^4 \times 0.3$ . Every other order with four makes, such as SFSSS, multiplies the same five numbers in a different order, so it has the same probability. There are $\binom{5}{4} = 5$ such orders, one for each choice of the missed shot. Every arrangement with $k$ successes has the same probability, $p^k(1 - p)^{n - k}$ , so the answer is that probability times the number of arrangements, $\binom{n}{k}$ .

Doing this for every $k$ from $0$ to $5$ gives the whole distribution:

Makes in 5 free throws when p = 0.7

The bars add to $1$ . They peak at $4$ , close to $np = 5 \times 0.7 = 3.5$ , the number of makes she averages over many sets of five shots.

Worked examples

Common mistakes

Practice problems

Find the probability of exactly $2$ heads in $4$ flips of a fair coin.

Answer

$0.375$

Full solution

$\binom{4}{2} = 6$ of the $16$ equally likely sequences have $2$ heads: $\tfrac{6}{16} = 0.375$ .
A die is rolled $4$ times. Find the probability of exactly one six.

Answer

About $0.386$

Full solution

$\binom{4}{1}\left(\tfrac{1}{6}\right)\left(\tfrac{5}{6}\right)^3 = 4 \times \tfrac{125}{1296} = \tfrac{500}{1296} \approx 0.386$ .
Each trial succeeds with probability $0.8$ . Find the probability that all $3$ of $3$ trials succeed.

Answer

$0.512$

Full solution

$\binom{3}{3}(0.8)^3(0.2)^0 = 0.512$ .
For the same trials, find the probability of at least one success.

Answer

$0.992$

Full solution

$1 - P(\text{none}) = 1 - 0.2^3 = 1 - 0.008$ .
$60\%$ of voters support a measure. Find the probability that exactly $3$ of $5$ randomly chosen voters support it.

Answer

$0.3456$

Full solution

$\binom{5}{3}(0.6)^3(0.4)^2 = 10 \times 0.216 \times 0.16 = 0.3456$ .
For the same voters, find the probability that at most $1$ of the $5$ supports it.

Answer

About $0.087$

Full solution

$P(0) + P(1) = 0.4^5 + 5(0.6)(0.4)^4 = 0.01024 + 0.0768 = 0.08704$ .
Two students are chosen without replacement from a class of $10$ with $4$ boys, and the boys are counted. Is the count binomial?

Answer

No

Full solution

The first choice is a boy with probability $\tfrac{4}{10}$ , but the second is a boy with probability $\tfrac{3}{9}$ or $\tfrac{4}{9}$ depending on the first. The probability changes, so the trials are not independent.
In how many ways can exactly $2$ successes occur in $6$ trials?

Answer

$15$

Full solution

Choose which $2$ of the $6$ trials succeed: $\binom{6}{2} = 15$ .
Seeds sprout with probability $0.9$ . A gardener plants $10$ . Find the probability that all $10$ sprout, and the probability that exactly $9$ sprout.

Answer

About $0.349$ and about $0.387$

Full solution

$0.9^{10} \approx 0.349$ . And $\binom{10}{9}(0.9)^9(0.1) = 10 \times 0.3874 \times 0.1 \approx 0.387$ . Exactly $9$ is more likely than all $10$ , because there are $10$ ways for one seed to fail.
A student finds the probability of exactly $2$ heads in $3$ flips as $(0.5)^2(0.5) = \tfrac{1}{8}$ . What went wrong?

Hint

List the sequences with two heads.

Answer

The student counted one order. There are $\binom{3}{2} = 3$ , so the probability is $\tfrac{3}{8}$ .

Full solution

HHT, HTH and THH each have probability $\tfrac{1}{8}$ , and together they give $\tfrac{3}{8}$ . The factor $\binom{3}{2}$ counts those orders.

Frequently asked questions

What is the binomial probability formula?

P(X = k) = C(n, k) · p^k · (1 − p)^(n − k), the probability of exactly k successes in n independent trials that each succeed with probability p.

What are the conditions for a binomial distribution?

Each trial has two outcomes, success or failure; the trials are independent; the number of trials n is fixed; and the probability of success p is the same on every trial.

Why is there a combination in the formula?

The k successes can fall on any k of the n trials. Each choice is a different outcome with the same probability, and the combination counts how many choices there are.

How do you find the probability of at least one success?

Use the complement: P(at least one) = 1 − P(none) = 1 − (1 − p)^n.

Is drawing cards without replacement binomial?

No. Each draw changes what is left in the deck, so the probability of success changes and the draws are not independent.

What to learn next

Random Variables and Expected ValueHow a random variable turns outcomes into numbers, how to build its probability distribution from theory or from data, and why expected value is the long-run average.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSS.CP.B.9Conditional Probability and the Rules of Probability(+) Use permutations and combinations to compute probabilities of compound events and solve problems.
CCSS.MATH.CONTENT.HSS.MD.A.3Using Probability to Make Decisions(+) Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value.

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.