Binomial Distribution

Learn how to calculate probabilities for repeated independent trials with two outcomes.

Advanced30 minLesson
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Definition

The binomial distribution models the number of successes in a fixed number of independent trials, where each trial has only two possible outcomes: success or failure.
The probability of exactly kk successes in nn trials is:
P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
Where:
  • nn = number of trials
  • kk = number of successes
  • pp = probability of success on each trial
  • (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!} = number of ways to choose kk successes from nn trials
Conditions for a Binomial Distribution:
  1. 1.Fixed number of trials (nn)
  2. 2.Each trial is independent
  3. 3.Only two outcomes per trial (success/failure)
  4. 4.Same probability of success (pp) for each trial

Try it now

Which of the following is a condition for a binomial distribution?

Worked Examples

You flip a fair coin 5 times. What is the probability of getting exactly 3 heads?
1

Identify the parameters

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

Calculate the combination

(53)=5!3!(5−3)!=5!3!2!=1206×2=10\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10 → 10 ways
3

Calculate success probability

pk=0.53=0.125p^k = 0.5^3 = 0.125
4

Calculate failure probability

(1−p)n−k=0.52=0.25(1-p)^{n-k} = 0.5^2 = 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 → 0.3125 or 31.25%

Common Mistakes

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

Why it's wrong: The combination counts HOW MANY WAYS to arrange kk successes among nn trials. Without it, you only calculate the probability of ONE specific arrangement.

Correct: Always use the full formula: P(X=k)=(nk)pk(1−p)n−kP(X=k) = \binom{n}{k} p^k (1-p)^{n-k}

Using pp for both success AND failure terms

Why it's wrong: The failure probability is 1−p1-p, not pp. If success is 0.3, failure must be 0.7.

Correct: Use pkp^k for successes and (1−p)n−k(1-p)^{n-k} for failures.

Confusing 'at least' with 'exactly'

Why it's wrong: 'Exactly 3' means P(X=3)P(X=3), but 'at least 3' means P(X≥3)=P(X=3)+P(X=4)+...P(X \geq 3) = P(X=3) + P(X=4) + ...

Correct: For 'at least', add up probabilities for all values from that number to nn. Or use complement: P(X≥k)=1−P(X<k)P(X \geq k) = 1 - P(X < k)

Interactive Visual

Bar Chart

n = 5, p = 0.5 (fair coin)

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

Bar Chart

n = 6, p = 0.75 (free throws)

0(0)
1(0)
2(0.03)
3(0.13)
4(0.3)
5(0.36)
6(0.18)

Interactive Sandbox

Expression Calculator

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Practice Problems

17 problems
Problem 1 of 17
Easy
Which of the following is a condition for a binomial distribution?

Why It Matters

The binomial distribution appears everywhere in real life:
  • Quality Control: What's the probability that 3 out of 100 products are defective?
  • Medicine: If a drug works 80% of the time, what's the chance it helps at least 7 out of 10 patients?
  • Sports: What's the probability a basketball player makes exactly 6 out of 10 free throws?
  • Genetics: If a trait has 25% inheritance probability, how likely is it that 2 out of 4 children inherit it?
Understanding binomial distribution helps you make predictions and informed decisions when dealing with repeated yes/no situations.

Real World Applications

Medical Trials

Pharmaceutical companies use binomial distribution to analyze drug effectiveness in clinical trials.

Example:

If a new medication has a 75% success rate, the probability that exactly 8 out of 10 patients improve is calculated using the binomial formula.

1Try It Yourself

A vaccine is 90% effective. You vaccinate 5 people.

What is the probability that exactly 4 people develop immunity?

Step 1: Write the mathematical expression

Use the binomial formula with n=5n=5, k=4k=4, p=0.9p=0.9:

Sports Analytics

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

Example:

If a basketball player has a 70% free throw rate, binomial distribution calculates the probability of making a specific number of shots.

2Try It Yourself

A soccer player scores penalty kicks 80% of the time. They take 4 penalties.

What is the probability they score exactly 3?

Step 1: Write the mathematical expression

Apply the binomial formula:

Key Takeaways

  • 1The binomial distribution models kk successes in nn independent trials with success probability pp
  • 2Formula: 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 the number of ways to arrange the successes
  • 4Conditions: fixed nn, independent trials, two outcomes, constant pp
  • 5Mean: μ=np\mu = np and Standard Deviation: σ=np(1−p)\sigma = \sqrt{np(1-p)}

Frequently Asked Questions

Use it when you have a fixed number of independent trials, each with only two outcomes (success/failure), and the probability of success remains constant for each trial.
Use it when you have a fixed number of independent trials, each with only two outcomes (success/failure), and the probability of success remains constant for each trial.
Binomial is for discrete counts (whole numbers of successes), while normal is for continuous data. However, when nn is large and pp is not too extreme, binomial can be approximated by normal.
Either add up P(X=k)+P(X=k+1)+...+P(X=n)P(X=k) + P(X=k+1) + ... + P(X=n), or use the complement: P(X≥k)=1−P(X<k)=1−[P(X=0)+P(X=1)+...+P(X=k−1)]P(X \geq k) = 1 - P(X < k) = 1 - [P(X=0) + P(X=1) + ... + P(X=k-1)].

Glossary

Binomial distribution
A probability distribution for the number of successes in a fixed number of independent trials
Trial
A single experiment or observation with two possible outcomes
Bernoulli trial
An experiment with exactly two outcomes: success (probability pp) or failure (probability 1−p1-p)
Combination
(nk)\binom{n}{k} represents the number of ways to choose kk items from nn items, regardless of order
Independent events
Events where the outcome of one does not affect the probability of another

Formula Card

Probability

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

Probability of exactly k successes in n trials

Combination

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

Number of ways to choose k from n

Mean

μ=np\mu = np

Expected number of successes

Variance

σ2=np(1−p)\sigma^2 = np(1-p)

Spread of the distribution

Standard Deviation

σ=np(1−p)\sigma = \sqrt{np(1-p)}

Square root of variance

Today's challengeOne problem, three tries

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