Binomial Probability

Learn to calculate probabilities when an experiment has exactly two outcomes and is repeated multiple times.

Advanced25 minLesson
Printable worksheet

Definition

Binomial probability describes the chance of getting exactly kk successes in nn independent trials, where each trial has only two outcomes (success or failure) with the same probability.
The binomial probability formula is:
P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
Where:
  • nn = total number of trials
  • kk = number of successes we want
  • pp = probability of success on each trial
  • (nk)\binom{n}{k} = number of ways to choose kk successes from nn trials (combinations)
The combination formula is:
(nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}

Try it now

Which of the following is a requirement for a binomial experiment?

Worked Examples

What is the probability of getting exactly 3 heads in 5 coin flips?
1

Identify the parameters

n=5n = 5 flips, k=3k = 3 heads, p=0.5p = 0.5 (fair coin) → n=5n=5, k=3k=3, p=0.5p=0.5
2

Calculate the combination

(53)=5!3!⋅2!=1206⋅2=10\binom{5}{3} = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \cdot 2} = 10 → (53)=10\binom{5}{3} = 10
3

Calculate pkp^k

(0.5)3=0.125(0.5)^3 = 0.125 → p3=0.125p^3 = 0.125
4

Calculate (1−p)n−k(1-p)^{n-k}

(0.5)5−3=(0.5)2=0.25(0.5)^{5-3} = (0.5)^2 = 0.25 → (1−p)2=0.25(1-p)^2 = 0.25
5

Multiply all parts

P(X=3)=10×0.125×0.25=0.3125P(X=3) = 10 \times 0.125 \times 0.25 = 0.3125 → P=0.3125P = 0.3125

Common Mistakes

Forgetting to include the combination coefficient (nk)\binom{n}{k}

Why it's wrong: The combination counts all the different ways to arrange successes among the trials. Without it, you only calculate one specific arrangement.

Correct: Always start with the combination: P(X=k)=(nk)×pk×(1−p)n−kP(X=k) = \binom{n}{k} \times p^k \times (1-p)^{n-k}

Using pnp^n instead of pkp^k

Why it's wrong: The exponent kk represents the number of successes, not the total number of trials.

Correct: Success probability is raised to the power of successes: pkp^k, and failure probability to the power of failures: (1−p)n−k(1-p)^{n-k}

Applying binomial formula when trials are not independent

Why it's wrong: Binomial probability requires that each trial's outcome doesn't affect others. Drawing cards without replacement, for example, changes probabilities.

Correct: Check independence: Does knowing one result change the probability of another? If yes, binomial doesn't apply.

Interactive Visual

Bar Chart

5 coin flips: P(exactly k heads)

0(0.03)
1(0.16)
2(0.31)
3(0.31)
4(0.16)
5(0.03)

Pascal's triangle

n=011n=1112n=21214n=313318n=41464116n=51510105132n=6161520156164

C(5, 3) = 10 = 6 + 4

Row 5: 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2⁵

Tap a number: it is the sum of the two numbers above it, and it equals C(n, k).

Interactive Sandbox

Expression Calculator

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Practice Problems

18 problems
Problem 1 of 18
Easy
Which of the following is a requirement for a binomial experiment?

Why It Matters

Binomial probability appears everywhere when outcomes are binary:
  • Medicine: Probability that exactly 7 out of 10 patients respond to a treatment
  • Quality Control: Chance of finding 3 defective items in a batch of 20
  • Sports: Likelihood of a basketball player making 8 out of 10 free throws
  • Surveys: Probability that exactly 6 of 10 respondents choose a particular option
Understanding binomial probability helps make predictions and informed decisions in uncertain situations!

Real World Applications

Medical Trials

Researchers use binomial probability to evaluate treatment effectiveness.

Example:

If a drug has a 70% success rate, we can calculate the probability that exactly 8 out of 10 patients will respond positively.

1Try It Yourself

A vaccine is 80% effective. In a group of 5 people who received it, what's the probability that exactly 4 are protected?

Calculate P(X=4)P(X = 4) where n=5n=5, p=0.8p=0.8

Step 1: Write the mathematical expression

Use: (54)×(0.8)4×(0.2)1\binom{5}{4} \times (0.8)^4 \times (0.2)^1

Sports Analytics

Coaches use binomial probability to predict game outcomes and player performance.

Example:

A basketball player has a 75% free throw success rate. Binomial probability helps calculate the chance of making 9 out of 10 free throws.

2Try It Yourself

A soccer player scores on 60% of penalty kicks. What's the probability of scoring exactly 3 out of 5 penalties?

Calculate P(X=3)P(X = 3) where n=5n=5, p=0.6p=0.6

Step 1: Write the mathematical expression

Use: (53)×(0.6)3×(0.4)2\binom{5}{3} \times (0.6)^3 \times (0.4)^2

Key Takeaways

  • 1Binomial probability applies when there are exactly two outcomes (success/failure) with fixed probability
  • 2The formula is P(X=k)=(nk)pk(1−p)n−kP(X=k) = \binom{n}{k} p^k (1-p)^{n-k}
  • 3The combination (nk)\binom{n}{k} counts all possible arrangements of successes
  • 4Trials must be independent for binomial probability to apply
  • 5Always check: fixed nn, constant pp, independent trials, two outcomes

Frequently Asked Questions

Use binomial probability when you have: (1) a fixed number of trials, (2) exactly two outcomes per trial, (3) constant probability of success, and (4) independent trials. If any condition fails, you need a different approach.
Use binomial probability when you have: (1) a fixed number of trials, (2) exactly two outcomes per trial, (3) constant probability of success, and (4) independent trials. If any condition fails, you need a different approach.
It counts how many different ways you can arrange kk successes among nn trials. For example, getting 2 heads in 3 flips can happen as HHT, HTH, or THH - that's (32)=3\binom{3}{2} = 3 ways.
For 'at least kk', add up P(X=k)+P(X=k+1)+...+P(X=n)P(X=k) + P(X=k+1) + ... + P(X=n). For 'at most kk', add up P(X=0)+P(X=1)+...+P(X=k)P(X=0) + P(X=1) + ... + P(X=k). Sometimes it's easier to use the complement.

Glossary

Binomial experiment
An experiment with a fixed number of independent trials, each with exactly two outcomes (success/failure) and constant probability
Trial
A single repetition of the experiment (e.g., one coin flip, one question answered)
Success
The outcome we're counting (doesn't have to be 'good' - just the one we're interested in)
Combination
The number of ways to choose kk items from nn items, written as (nk)\binom{n}{k} or C(n,k)C(n,k)
Independent trials
Trials where the outcome of one doesn't affect the probability of others

Formula Card

Binomial Probability Formula

P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

$n$ = trials, $k$ = successes, $p$ = success probability, $\binom{n}{k}$ = combinations

Combination Formula

(nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}

Number of ways to choose $k$ items from $n$ items

Today's challengeOne problem, three tries

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