Binomial Probability Calculator
Statistics & probability • 2026 edition
Binomial Distribution Formula:
Show the calculator\( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
Where:
- \( P(X = k) \) = Probability of exactly k successes
- \( n \) = Number of trials
- \( k \) = Number of successes
- \( p \) = Probability of success on each trial
- \( \binom{n}{k} \) = Binomial coefficient = \( \frac{n!}{k!(n-k)!} \)
The binomial distribution models the number of successes in a fixed number of independent Bernoulli trials with the same probability of success. It's fundamental for modeling discrete events with binary outcomes like coin flips, pass/fail tests, or yes/no responses.
Example: For n=10, k=3, p=0.2:
\( \binom{10}{3} = \frac{10!}{3!7!} = 120 \)
\( P(X = 3) = 120 \times (0.2)^3 \times (0.8)^7 = 120 \times 0.008 \times 0.2097152 = 0.2013 \)
So there's a 20.13% chance of exactly 3 successes in 10 trials.
Input Parameters
Advanced Options
Results
Comprehensive Statistics Guide
The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials with the same probability of success. It's characterized by two parameters: n (number of trials) and p (probability of success). The distribution is fundamental for modeling binary outcomes in repeated experiments.
The probability mass function:
Where the binomial coefficient is:
Key parameters of binomial distribution:
- Mean: μ = np
- Variance: σ² = np(1-p)
- Standard Deviation: σ = √[np(1-p)]
- Mode: ⌊(n+1)p⌋ (when not integer) or both ⌊(n+1)p⌋ and ⌊(n+1)p⌋-1
- Quality Control: Defective items in batches
- Medical Trials: Success rates of treatments
- Marketing: Response rates to campaigns
- Finance: Default rates in portfolios
- Education: Pass/fail rates in exams
Statistics Fundamentals
Discrete distribution modeling number of successes in n independent trials with probability p.
\( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
Where n=trials, k=successes, p=probability of success.
- k must be between 0 and n
- 0 ≤ p ≤ 1
- Trials must be independent
Applications
Used in quality control, medical research, finance, marketing, and experimental design.
- Verify binomial conditions
- Identify n, k, and p
- Apply appropriate formula
- Calculate exact or cumulative probability
- Interpret results in context
- Verify assumptions
- Check independence assumption
- Verify constant probability
- Consider sample size limitations
- Account for rare events
Statistics Learning Quiz
Which of the following scenarios can be modeled using a binomial distribution?
The answer is B) Flipping a fair coin 10 times and counting heads. This scenario satisfies all requirements for a binomial distribution: fixed number of trials (n=10), binary outcomes (heads/tails), constant probability of success (p=0.5), and independent trials.
For a binomial distribution, four conditions must be met: 1) Fixed number of trials, 2) Binary outcomes for each trial, 3) Constant probability of success, 4) Independent trials. Drawing cards without replacement violates the independence condition, measuring heights is continuous, and time-to-event is modeled by other distributions.
Binomial Distribution: Discrete distribution for count of successes
Independence: Outcome of one trial doesn't affect others
Binary Outcomes: Only two possible outcomes per trial
• Must have fixed number of trials
• Each trial must be independent
• Probability must remain constant
• Look for "number of successes in n trials"
• Verify independence assumption
• Check for binary outcomes
• Applying to non-independent trials
• Using with continuous outcomes
• Forgetting to verify conditions
A basketball player makes 75% of his free throws. If he shoots 4 free throws, what is the probability he makes exactly 3? Show your work.
Given: n = 4 (trials), k = 3 (successes), p = 0.75 (probability of success)
Step 1: Calculate binomial coefficient:
\( \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3! \times 1!} = \frac{24}{6 \times 1} = 4 \)
Step 2: Calculate probability terms:
\( p^k = (0.75)^3 = 0.421875 \)
\( (1-p)^{n-k} = (0.25)^1 = 0.25 \)
Step 3: Apply binomial formula:
\( P(X = 3) = \binom{4}{3} \times (0.75)^3 \times (0.25)^1 = 4 \times 0.421875 \times 0.25 = 0.421875 \)
Therefore, there is a 42.19% probability the player makes exactly 3 out of 4 free throws.
This problem demonstrates the systematic approach to binomial calculations. First, identify the parameters (n, k, p). Then calculate the binomial coefficient, followed by the probability terms, and finally multiply them together. The binomial coefficient counts the number of ways to arrange the successes among the trials.
Binomial Coefficient: Number of ways to choose k successes from n trials
Success Probability: Probability of success on each trial
Exact Probability: Probability of exactly k successes
• Calculate binomial coefficient first
• Raise probabilities to appropriate powers
• Multiply all terms together
• Use calculator for binomial coefficients
• Remember: n! = n × (n-1)!
• Check that probability is between 0 and 1
• Incorrectly calculating binomial coefficient
• Using wrong exponents
• Forgetting to multiply all terms
A manufacturing process produces 5% defective items. If a quality inspector examines 20 items, what is the probability that at most 2 items are defective? What is the expected number of defective items?
Given: n = 20, p = 0.05, need P(X ≤ 2)
Step 1: Calculate P(X = 0), P(X = 1), and P(X = 2)
P(X = 0) = C(20,0) × (0.05)⁰ × (0.95)²⁰ = 1 × 1 × 0.3585 = 0.3585
P(X = 1) = C(20,1) × (0.05)¹ × (0.95)¹⁹ = 20 × 0.05 × 0.3774 = 0.3774
P(X = 2) = C(20,2) × (0.05)² × (0.95)¹⁸ = 190 × 0.0025 × 0.3972 = 0.1887
Step 2: Sum the probabilities:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.3585 + 0.3774 + 0.1887 = 0.9246
Step 3: Calculate expected value:
Mean = np = 20 × 0.05 = 1.0
Therefore, there is a 92.46% probability of at most 2 defective items, and the expected number of defective items is 1.0.
This problem combines exact and cumulative probabilities. For "at most 2", we sum P(X=0) + P(X=1) + P(X=2). The expected value formula μ=np is straightforward and represents the long-run average number of successes. This is a common quality control application.
Cumulative Probability: Probability of at most k successes
Expected Value: Long-run average number of successes
Quality Control: Statistical methods for monitoring processes
• P(X ≤ k) = P(X = 0) + P(X = 1) + ... + P(X = k)
• Mean = np for binomial distribution
• Variance = np(1-p)
• Use symmetry for calculations when possible
• Remember mean formula for quick estimates
• Check that probabilities sum to reasonable value
• Forgetting to sum all required terms
• Misinterpreting "at most" vs "exactly"
• Using wrong formula for expected value
A new drug is successful in treating a disease 60% of the time. If 15 patients receive the drug, what is the probability that more than 10 patients recover? What is the probability that between 8 and 12 patients (inclusive) recover?
Given: n = 15, p = 0.6
Part A: P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)
Calculating each term:
P(X = 11) = C(15,11) × (0.6)¹¹ × (0.4)⁴ ≈ 0.1268
P(X = 12) = C(15,12) × (0.6)¹² × (0.4)³ ≈ 0.0634
P(X = 13) = C(15,13) × (0.6)¹³ × (0.4)² ≈ 0.0219
P(X = 14) = C(15,14) × (0.6)¹⁴ × (0.4)¹ ≈ 0.0047
P(X = 15) = C(15,15) × (0.6)¹⁵ × (0.4)⁰ ≈ 0.0005
P(X > 10) ≈ 0.1268 + 0.0634 + 0.0219 + 0.0047 + 0.0005 = 0.2173
Part B: P(8 ≤ X ≤ 12) = P(X ≤ 12) - P(X ≤ 7)
Using cumulative probabilities: P(X ≤ 12) ≈ 0.9730, P(X ≤ 7) ≈ 0.2131
P(8 ≤ X ≤ 12) = 0.9730 - 0.2131 = 0.7599
Therefore, there is a 21.73% probability that more than 10 patients recover, and a 75.99% probability that between 8 and 12 patients recover.
This problem demonstrates calculating probabilities for ranges. For "more than 10", we sum P(X=11) through P(X=15). For "between 8 and 12", we use the cumulative approach: P(X≤12) - P(X≤7). This is more efficient than summing individual probabilities for each value in the range.
Range Probability: Probability of success count within a range
Cumulative Approach: Using CDF to find range probabilities
Medical Success Rate: Probability of treatment effectiveness
• P(a ≤ X ≤ b) = P(X ≤ b) - P(X ≤ a-1)
• P(X > k) = 1 - P(X ≤ k)
• P(X ≥ k) = 1 - P(X ≤ k-1)
• Use complement for "greater than" problems
• Use cumulative approach for ranges
• Consider normal approximation for large n
• Forgetting to subtract P(X ≤ a-1) for ranges
• Miscounting the number of terms to sum
• Not considering the complement approach
For a binomial distribution with n = 20 and p = 0.3, which of the following statements is TRUE?
The answer is C) Mean = 6. For a binomial distribution, the mean is μ = np = 20 × 0.3 = 6. The variance is σ² = np(1-p) = 20 × 0.3 × 0.7 = 4.2, and the standard deviation is σ = √4.2 ≈ 2.05.
This question tests knowledge of binomial distribution parameters. The mean is always np, the variance is always np(1-p), and the standard deviation is the square root of the variance. These formulas are fundamental to understanding the distribution's behavior.
Mean: Expected value of the distribution
Variance: Measure of spread around the mean
Standard Deviation: Square root of variance
• Mean = np
• Variance = np(1-p)
• Std Dev = √[np(1-p)]
• Remember the mean formula for quick estimates
• Variance is always smaller than mean when p < 0.5
• Check that standard deviation is reasonable
• Confusing mean and variance formulas
• Forgetting to take square root for std dev
• Using wrong values for calculations
FAQ
Q: When can I use normal approximation to binomial distribution?
A: The normal approximation is appropriate when both np ≥ 5 and n(1-p) ≥ 5. This ensures sufficient sample size in both success and failure categories. The approximation uses:
\( X \sim N(np, np(1-p)) \)
For example, with n=100 and p=0.1: np = 10 ≥ 5 and n(1-p) = 90 ≥ 5, so normal approximation is suitable.
Apply continuity correction: P(X ≤ k) ≈ P(X ≤ k + 0.5) with normal distribution.
Q: What's the difference between binomial and geometric distribution?
A: Key differences:
Binomial: Fixed number of trials n, variable number of successes
\( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
Geometric: Variable number of trials until first success, fixed number of successes (1)
\( P(X = k) = p(1-p)^{k-1} \)
Binomial answers "How many successes in n trials?" Geometric answers "How many trials until first success?"