Binomial Distribution Calculator
Calculate exact and cumulative binomial probabilities for n trials with success probability p. Explore the distribution with interactive charts and key statistics.
Calculate exact and cumulative binomial probabilities for n trials with success probability p. Explore the distribution with interactive charts and key statistics.
The binomial distribution is one of the most fundamental discrete probability distributions in statistics, modeling the number of successes in a fixed number of independent trials where each trial has exactly two possible outcomes (success or failure) and the probability of success remains constant. This tool helps you calculate exact probabilities P(X = k), cumulative probabilities P(X ≤ k) and P(X ≥ k), and range probabilities P(a ≤ X ≤ b) for binomial experiments. Whether you're a student learning probability theory, a researcher analyzing experimental data, a quality control engineer monitoring defect rates, or a business professional making data-driven decisions, understanding binomial distributions enables you to quantify uncertainty, make predictions, and assess the likelihood of specific outcomes in binary experiments.
For students and researchers, this tool demonstrates practical applications of probability theory, combinatorics, and statistical modeling. The binomial distribution calculation shows how combinations (n choose k), probability multiplication, and cumulative sums combine to produce meaningful probability assessments. Students can use this tool to verify homework calculations, understand how parameters n (number of trials) and p (success probability) affect probability distributions, and explore concepts like mean, variance, and standard deviation in discrete probability contexts. Researchers can apply binomial distributions to analyze experimental data, test hypotheses about success rates, and model binary outcomes in fields ranging from medicine and biology to engineering and social sciences.
For business professionals and practitioners, binomial distributions provide essential tools for decision-making under uncertainty. Quality control engineers use binomial models to assess defect rates in manufacturing, calculate acceptance probabilities for sampling plans, and determine whether production processes meet quality standards. Marketing professionals use binomial distributions to model conversion rates, analyze A/B test results, and predict customer behavior. Medical researchers use binomial models to analyze clinical trial outcomes, assess treatment effectiveness, and calculate confidence intervals for success rates. Financial analysts use binomial models to price options, assess investment risks, and model binary financial outcomes.
For the common person, this tool answers practical probability questions: What's the chance of getting exactly 7 heads when flipping a coin 10 times? What's the probability of passing a multiple-choice test by guessing? What's the likelihood that at least 3 out of 5 products are defective? The tool calculates exact probabilities (P(X = k)), cumulative probabilities up to a value (P(X ≤ k)), cumulative probabilities from a value (P(X ≥ k)), and range probabilities (P(a ≤ X ≤ b)), providing comprehensive probability assessments for any binomial scenario. Taxpayers and budget-conscious individuals can use binomial distributions to model financial outcomes, assess risk in decision-making, and understand probability in everyday situations like lottery odds, insurance claims, and investment returns.
A binomial experiment consists of n independent trials, where each trial has exactly two possible outcomes: success (with probability p) or failure (with probability 1-p). The key characteristics are: (1) Fixed number of trials n, (2) Each trial is independent—the outcome of one trial doesn't affect another, (3) Two outcomes only—each trial results in either success or failure, (4) Constant probability—the probability p of success remains the same for every trial. Examples include flipping a coin n times (success = heads, p = 0.5), testing n products for defects (success = defective, p = defect rate), or surveying n people (success = yes answer, p = proportion who say yes).
The probability mass function P(X = k) gives the probability of getting exactly k successes in n trials. The formula is: P(X = k) = C(n, k) × p^k × (1-p)^(n-k), where C(n, k) is the binomial coefficient "n choose k" representing the number of ways to choose k successes from n trials. The binomial coefficient equals n! / (k! × (n-k)!) and counts the number of different sequences that result in exactly k successes. The term p^k represents the probability of k successes, while (1-p)^(n-k) represents the probability of (n-k) failures. Multiplying these terms gives the probability of any specific sequence with k successes, and multiplying by C(n, k) accounts for all possible sequences.
The cumulative distribution function P(X ≤ k) gives the probability of getting k or fewer successes in n trials. It's calculated by summing the PMF values from 0 to k: P(X ≤ k) = Σ(i=0 to k) P(X = i). This cumulative probability answers "at most" or "no more than" questions. For example, P(X ≤ 7) is the probability of getting 7 or fewer successes. The complementary cumulative probability P(X ≥ k) gives the probability of getting k or more successes, calculated as P(X ≥ k) = 1 - P(X ≤ k-1) = 1 - CDF(k-1). This answers "at least" or "no fewer than" questions. For example, P(X ≥ 3) is the probability of getting 3 or more successes.
The mean (expected value) of a binomial distribution is μ = n × p, representing the average number of successes you expect over many repetitions of the experiment. For example, with n=100 trials and p=0.5, you expect about 50 successes on average. The variance is σ² = n × p × (1-p), measuring how spread out the distribution is from the mean. The standard deviation is σ = √(n × p × (1-p)), which is easier to interpret since it's in the same units as the number of successes. About 68% of outcomes fall within one standard deviation of the mean, and about 95% fall within two standard deviations. The variance is maximized when p = 0.5 (most uncertainty) and minimized when p approaches 0 or 1 (least uncertainty).
The binomial coefficient C(n, k) = n! / (k! × (n-k)!) counts the number of ways to choose k successes from n trials, or equivalently, the number of different sequences that result in exactly k successes. For example, C(5, 2) = 10 means there are 10 different ways to get exactly 2 successes in 5 trials. The coefficient is calculated using a multiplicative formula for numerical stability: C(n, k) = (n × (n-1) × ... × (n-k+1)) / (k × (k-1) × ... × 1). This formula avoids computing large factorials and is more efficient for large n. The coefficient is symmetric: C(n, k) = C(n, n-k), which reflects that choosing k successes is equivalent to choosing (n-k) failures.
The tool supports four types of probability queries: (1) Exact Probability P(X = k)—the probability of getting exactly k successes, useful for specific outcome questions. (2) Cumulative Up To P(X ≤ k)—the probability of getting k or fewer successes, useful for "at most" or "no more than" questions. (3) Cumulative From P(X ≥ k)—the probability of getting k or more successes, useful for "at least" or "no fewer than" questions. (4) Range Probability P(a ≤ X ≤ b)—the probability that the number of successes falls between a and b (inclusive), calculated as P(X ≤ b) - P(X ≤ a-1), useful for interval questions. Each query type answers different practical questions and requires different calculations.
The shape of a binomial distribution depends on the success probability p. When p = 0.5, the distribution is symmetric around the mean n/2. When p < 0.5, the distribution is right-skewed (more probability mass on lower values, tail extends to the right). When p > 0.5, the distribution is left-skewed (more probability mass on higher values, tail extends to the left). The skewness becomes more pronounced as p moves further from 0.5. For large n, the binomial distribution approaches a normal distribution (by the Central Limit Theorem), making normal approximation useful when n is large and p is not too close to 0 or 1.
Start by entering the number of trials n in the "Number of Trials" field. This is the fixed number of independent experiments you're conducting. For example, if you're flipping a coin 10 times, n = 10. If you're testing 50 products for defects, n = 50. The tool accepts n values from 0 to 200. For very large n (>200), consider using normal approximation instead of exact binomial calculations. Make sure n represents the total number of independent trials in your experiment.
Enter the probability of success p in the "Success Probability" field. This is the probability that any single trial results in a success, and must be between 0 and 1 (inclusive). You can enter p as a decimal (e.g., 0.5 for 50%) or as a percentage (e.g., 50 for 50%). For example, if flipping a fair coin, p = 0.5. If 20% of products are typically defective, p = 0.2. If you're guessing on a multiple-choice question with 4 choices, p = 0.25. The tool will automatically convert percentages to decimals if needed.
Choose the type of probability you want to calculate: "Exact" for P(X = k), "Cumulative Up To" for P(X ≤ k), "Cumulative From" for P(X ≥ k), or "Between" for P(a ≤ X ≤ b). Select the option that matches your question. For example, if you want to know the probability of exactly 7 successes, choose "Exact". If you want to know the probability of at most 7 successes, choose "Cumulative Up To". If you want to know the probability of at least 7 successes, choose "Cumulative From". If you want to know the probability of between 5 and 7 successes, choose "Between".
Depending on your selected query type, enter the appropriate value(s). For "Exact", "Cumulative Up To", or "Cumulative From", enter the value x (the number of successes you're interested in). For "Between", enter the range start (a) and range end (b) values. Make sure x is between 0 and n (inclusive), and that range start ≤ range end and both are between 0 and n. The tool will validate your inputs and show an error if values are out of range or invalid.
Click "Calculate" or submit the form to compute the requested probability. The tool displays the calculated probability (as both a decimal and percentage), key statistics (mean, variance, standard deviation), a complete distribution table showing probabilities for all possible values of k from 0 to n, and an interactive chart visualizing the probability distribution. The interpretation summary explains what the probability means in practical terms, helping you understand the result in context.
Review the distribution table to see probabilities for all possible outcomes (k = 0 to n). Each row shows k (number of successes), the exact probability P(X = k), and the cumulative probability P(X ≤ k). The chart visualizes the probability mass function, showing how probability is distributed across different values of k. Use the table and chart to understand the shape of the distribution, identify the most likely outcomes (highest probabilities), and see how probabilities change as k increases. The mean, variance, and standard deviation provide summary statistics that characterize the distribution's center and spread.
The binomial coefficient C(n, k) is calculated using a multiplicative formula for numerical stability:
If k < 0 or k > n: C(n, k) = 0
If k = 0 or k = n: C(n, k) = 1
Otherwise: Use k = min(k, n-k) for efficiency
C(n, k) = (n × (n-1) × ... × (n-k+1)) / (k × (k-1) × ... × 1)
This multiplicative formula avoids computing large factorials and is more numerically stable than the factorial formula n! / (k! × (n-k)!). The formula multiplies k terms in the numerator (n down to n-k+1) and divides by k terms in the denominator (k down to 1), producing the same result as the factorial formula but with better numerical properties.
The PMF P(X = k) is calculated using the binomial probability formula:
If k < 0 or k > n: P(X = k) = 0
If p = 0: P(X = k) = 1 if k = 0, else 0
If p = 1: P(X = k) = 1 if k = n, else 0
Otherwise: P(X = k) = C(n, k) × p^k × (1-p)^(n-k)
The formula multiplies three components: the binomial coefficient C(n, k) (number of ways to get k successes), p^k (probability of k successes), and (1-p)^(n-k) (probability of n-k failures). The result is clamped to [0, 1] to handle floating-point errors. Special cases (p = 0 or p = 1) are handled separately to avoid numerical issues.
The CDF P(X ≤ k) is calculated by summing PMF values from 0 to k:
If k < 0: P(X ≤ k) = 0
If k ≥ n: P(X ≤ k) = 1
Otherwise: P(X ≤ k) = Σ(i=0 to k) P(X = i)
The CDF is computed by iterating from i = 0 to k, calculating each PMF value P(X = i), and summing them. The result is clamped to [0, 1] to handle floating-point errors. For efficiency, the calculation can be optimized using recurrence relations or specialized algorithms, but the direct summation method is used here for clarity and accuracy.
The complementary cumulative probability P(X ≥ k) is calculated using the CDF:
P(X ≥ k) = 1 - P(X ≤ k-1) = 1 - CDF(k-1)
This formula uses the complement rule: the probability of getting k or more successes equals 1 minus the probability of getting k-1 or fewer successes. This is more efficient than summing PMF values from k to n, especially when k is large. The formula handles edge cases: if k = 0, P(X ≥ 0) = 1 (always true); if k > n, P(X ≥ k) = 0 (impossible).
The range probability P(a ≤ X ≤ b) is calculated using CDF values:
P(a ≤ X ≤ b) = P(X ≤ b) - P(X ≤ a-1) = CDF(b) - CDF(a-1)
This formula uses the fact that P(a ≤ X ≤ b) equals the probability of getting b or fewer successes minus the probability of getting (a-1) or fewer successes. This gives the probability of all outcomes from a to b inclusive. The formula handles edge cases: if a = 0, P(0 ≤ X ≤ b) = CDF(b); if a > b, the probability is 0 (invalid range).
The mean, variance, and standard deviation are calculated using standard formulas:
Mean (Expected Value): μ = n × p
Variance: σ² = n × p × (1-p)
Standard Deviation: σ = √(n × p × (1-p))
These formulas are derived from the properties of the binomial distribution and don't require summing over all possible outcomes. The mean represents the expected number of successes, the variance measures spread around the mean, and the standard deviation provides an interpretable measure of variability in the same units as the number of successes.
Let's calculate the probability of getting exactly 7 heads when flipping a fair coin 10 times:
Given: n = 10, p = 0.5, k = 7
Step 1: Calculate Binomial Coefficient C(10, 7)
C(10, 7) = C(10, 3) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120
Step 2: Calculate PMF P(X = 7)
P(X = 7) = C(10, 7) × (0.5)^7 × (0.5)^(10-7)
= 120 × (0.5)^7 × (0.5)^3
= 120 × 0.0078125 × 0.125
= 120 × 0.0009765625 = 0.1171875 ≈ 0.1172 (11.72%)
Step 3: Calculate Summary Statistics
Mean: μ = 10 × 0.5 = 5.0
Variance: σ² = 10 × 0.5 × 0.5 = 2.5
Standard Deviation: σ = √2.5 ≈ 1.58
Interpretation:
The probability of getting exactly 7 heads in 10 coin flips is approximately 11.72%. You expect 5 heads on average, with a standard deviation of about 1.58 heads.
Now let's calculate the cumulative probability P(X ≤ 7):
Calculate P(X ≤ 7)
P(X ≤ 7) = P(X = 0) + P(X = 1) + ... + P(X = 7)
Using the PMF formula for each value and summing:
P(X ≤ 7) ≈ 0.9453 (94.53%)
Interpretation:
The probability of getting 7 or fewer heads in 10 coin flips is approximately 94.53%. This means there's a 94.53% chance you'll get at most 7 heads.
This example demonstrates how the binomial distribution models the probability of successes in fixed trials. The exact probability P(X = 7) = 11.72% is relatively low because getting exactly 7 heads is a specific outcome, while the cumulative probability P(X ≤ 7) = 94.53% is much higher because it includes all outcomes from 0 to 7.
A student takes a 20-question multiple-choice test with 4 choices per question and guesses on all questions. They need at least 10 correct answers to pass. Using the tool with n=20, p=0.25 (1/4 chance of guessing correctly), and "Cumulative From" mode with x=10, the tool calculates P(X ≥ 10) ≈ 0.0139 (1.39%). The student learns that the probability of passing by guessing alone is very low (1.39%), demonstrating why studying is important. The mean is 5 (expected correct answers), variance is 3.75, and standard deviation is about 1.94, showing that most outcomes cluster around 5 correct answers.
A quality control engineer tests 50 products from a production line where the historical defect rate is 5%. They want to know the probability of finding at most 2 defective products. Using the tool with n=50, p=0.05, and "Cumulative Up To" mode with x=2, the tool calculates P(X ≤ 2) ≈ 0.5405 (54.05%). The engineer learns that there's a 54.05% chance of finding 2 or fewer defects, which helps assess whether the production process is meeting quality standards. The mean is 2.5 (expected defects), variance is 2.375, and standard deviation is about 1.54.
A medical researcher conducts a clinical trial with 100 patients where the treatment has a 60% success rate. They want to know the probability that at least 70 patients respond positively. Using the tool with n=100, p=0.60, and "Cumulative From" mode with x=70, the tool calculates P(X ≥ 70) ≈ 0.0162 (1.62%). The researcher learns that the probability of 70 or more successes is relatively low (1.62%), which helps assess treatment effectiveness and determine if results are statistically significant. The mean is 60 (expected successes), variance is 24, and standard deviation is about 4.90.
A person flips a fair coin 10 times and wants to know the probability of getting exactly 7 heads. Using the tool with n=10, p=0.5, and "Exact" mode with x=7, the tool calculates P(X = 7) ≈ 0.1172 (11.72%). The person learns that getting exactly 7 heads has an 11.72% probability, which is relatively low because it's a specific outcome. They can also check the distribution table to see probabilities for all outcomes (0 to 10 heads) and understand that 5 heads is the most likely outcome (mean = 5.0).
A marketing professional runs a campaign targeting 200 potential customers where the historical conversion rate is 8%. They want to know the probability that between 15 and 20 customers will convert. Using the tool with n=200, p=0.08, and "Between" mode with rangeStart=15 and rangeEnd=20, the tool calculates P(15 ≤ X ≤ 20) ≈ 0.4234 (42.34%). The professional learns that there's a 42.34% chance of getting 15 to 20 conversions, which helps set realistic expectations and assess campaign performance. The mean is 16 (expected conversions), variance is 14.72, and standard deviation is about 3.84.
A researcher surveys 150 people where 40% typically answer "yes" to a question. They want to know the probability that at most 50 people answer "yes". Using the tool with n=150, p=0.40, and "Cumulative Up To" mode with x=50, the tool calculates P(X ≤ 50) ≈ 0.0123 (1.23%). The researcher learns that the probability of 50 or fewer "yes" answers is very low (1.23%), which suggests that if they observe 50 or fewer "yes" answers, it might indicate a change in the population or sampling issue. The mean is 60 (expected "yes" answers), variance is 36, and standard deviation is 6.
A user wants to understand how changing n and p affects probabilities. They compare three scenarios: (1) n=10, p=0.5, x=5 gives P(X=5) ≈ 0.2461 (24.61%), (2) n=20, p=0.5, x=10 gives P(X=10) ≈ 0.1762 (17.62%), (3) n=10, p=0.3, x=3 gives P(X=3) ≈ 0.2668 (26.68%). The user learns that doubling n while keeping p constant reduces the exact probability (because there are more possible outcomes), and changing p shifts the distribution (lower p means lower mean and different probability distribution). This helps them understand how binomial parameters affect probability calculations.
The binomial distribution requires that trials are independent—the outcome of one trial must not affect another. Don't use binomial distributions for dependent trials, such as drawing cards without replacement (each draw affects the next), sampling from small populations without replacement, or situations where success in one trial influences the next. If trials are dependent, use other distributions like hypergeometric (for sampling without replacement) or consider the dependence structure in your model.
The binomial distribution assumes that the success probability p remains constant across all trials. Don't use binomial distributions when p varies between trials, such as when testing products from different production batches with different defect rates, when success probability changes over time, or when different trials have different conditions. If p varies, consider using other models or segmenting your analysis by groups with constant p.
P(X = k) is the probability of exactly k successes, while P(X ≤ k) is the probability of k or fewer successes (cumulative). Don't confuse these—P(X ≤ k) is always greater than or equal to P(X = k) because it includes all outcomes from 0 to k. For example, P(X = 7) might be 0.1172, while P(X ≤ 7) might be 0.9453. Make sure you select the correct query type ("Exact" vs "Cumulative Up To") based on whether your question asks for "exactly" or "at most".
The binomial distribution models experiments with exactly two outcomes (success or failure). Don't use binomial distributions for experiments with more than two outcomes, such as rolling a die (6 outcomes), selecting from multiple categories, or measuring continuous variables. For multiple outcomes, use multinomial distributions. For continuous variables, use appropriate continuous distributions like normal, exponential, or others.
The mean μ = n × p is the expected value (average over many repetitions), not necessarily the most likely single outcome. Don't assume that the mean is always the outcome with the highest probability. For example, with n=10 and p=0.5, the mean is 5, and P(X=5) is indeed the highest. But with n=10 and p=0.1, the mean is 1, and P(X=1) might not be the highest (P(X=0) or P(X=2) might be higher). Check the distribution table to find the mode (most likely outcome), which may differ from the mean.
For very large n (>200), exact binomial calculations can be slow and may have numerical precision issues. Don't use exact binomial calculations when n is extremely large—consider using normal approximation instead. The Central Limit Theorem states that for large n, the binomial distribution approaches a normal distribution with mean n×p and variance n×p×(1-p). Normal approximation is appropriate when n×p ≥ 5 and n×(1-p) ≥ 5, and provides good approximations for large n.
Make sure your inputs are within valid ranges: n must be a non-negative integer (0 to 200 in this tool), p must be between 0 and 1 (inclusive), and x (or range values) must be between 0 and n. Don't enter negative values, p > 1, x > n, or invalid ranges (rangeStart > rangeEnd). The tool validates inputs, but understanding valid ranges helps you avoid errors and interpret results correctly. Invalid inputs will produce errors or meaningless results.
After calculating a specific probability, review the complete distribution table to see probabilities for all possible outcomes (k = 0 to n). This helps you understand the shape of the distribution, identify the most likely outcomes (highest probabilities), see how probabilities change as k increases, and understand the relationship between exact and cumulative probabilities. The table provides context for your specific calculation and helps you interpret results in the broader distribution.
Use the mean μ = n × p to understand the expected number of successes, and the standard deviation σ = √(n × p × (1-p)) to understand typical variability. About 68% of outcomes fall within one standard deviation of the mean (μ ± σ), and about 95% fall within two standard deviations (μ ± 2σ). This helps you understand not just the expected value, but also the range of likely outcomes. For example, if μ = 50 and σ = 5, you expect most outcomes to fall between 45 and 55.
When your question involves a range of outcomes (e.g., "between 5 and 7 successes"), use the "Between" mode to calculate P(a ≤ X ≤ b) directly. This is more efficient than calculating multiple exact probabilities and summing them, and it's more accurate than approximations. Range probabilities are useful for questions like "What's the probability of getting 5 to 7 heads?" or "What's the chance that 10 to 15 products are defective?"
The shape of the binomial distribution depends on p: when p = 0.5, the distribution is symmetric; when p < 0.5, it's right-skewed (more probability on lower values); when p > 0.5, it's left-skewed (more probability on higher values). Understanding the shape helps you interpret probabilities—for right-skewed distributions, probabilities are higher for lower values of k, while for left-skewed distributions, probabilities are higher for higher values of k. Use the chart visualization to see the distribution shape.
Use complementary probabilities to verify calculations: P(X ≥ k) = 1 - P(X ≤ k-1) and P(X ≤ k) = 1 - P(X ≥ k+1). For example, if you calculate P(X ≤ 7) = 0.9453, you can verify by calculating P(X ≥ 8) = 1 - P(X ≤ 7) = 1 - 0.9453 = 0.0547, and checking that P(X ≤ 7) + P(X ≥ 8) = 1. This helps catch calculation errors and ensures probabilities sum correctly.
For large n (typically n > 200 or when n×p ≥ 5 and n×(1-p) ≥ 5), consider using normal approximation instead of exact binomial calculations. The normal approximation uses mean μ = n×p and variance σ² = n×p×(1-p), and provides good approximations when n is large and p is not too close to 0 or 1. Normal approximation is faster and avoids numerical precision issues with very large n, but exact binomial calculations are more accurate for small to moderate n.
Read the interpretation summary provided by the tool, which explains what your calculated probability means in practical terms. The summary includes the mean and standard deviation, explains the calculated probability in context, and provides insights about distribution shape (symmetric, right-skewed, or left-skewed). This helps you understand not just the number, but what it means for your specific scenario and how to use it in decision-making.
• Fixed Trial Count Required: The binomial distribution requires a predetermined, fixed number of trials (n). It cannot model situations where the number of trials is variable, unknown, or theoretically infinite—use Poisson or negative binomial distributions for such scenarios.
• Independence Assumption: Each trial must be completely independent of all other trials. The outcome of one trial cannot influence the outcome of any other trial. Violations occur in sampling without replacement from finite populations, sequential testing with adaptive rules, or processes where success changes future probabilities.
• Constant Success Probability: The probability of success (p) must remain identical across all trials. Real-world scenarios often have time-varying, condition-dependent, or learning-influenced success rates that violate this assumption and require more complex models.
• Numerical Precision Limits: For very large n (>170) or extreme probability values (p very close to 0 or 1), floating-point arithmetic may introduce small errors. Extremely small probabilities (below 10⁻¹⁵) may underflow to zero.
Important Note: This calculator is strictly for educational and informational purposes only. It does not provide professional statistical consulting, research validation, or scientific conclusions. The binomial model is an idealization—real-world phenomena rarely satisfy all assumptions perfectly. Results should be verified using professional statistical software (R, Python SciPy, SAS, SPSS, MATLAB) for any research, academic, quality control, or professional applications. For critical decisions in manufacturing, clinical trials, regulatory submissions, or financial risk assessment, always consult qualified statisticians who can evaluate assumption validity and recommend appropriate analytical approaches.
The mathematical formulas and statistical concepts used in this calculator are based on established probability theory and authoritative academic sources:
Common questions about binomial distributions, probability calculations, PMF and CDF formulas, assumptions, and how to use this calculator for homework and statistics practice.
The binomial distribution is used to model the number of successes in a fixed number of independent trials where each trial has only two possible outcomes (success or failure). Common applications include quality control (counting defects), clinical trials (counting positive responses), survey analysis (counting 'yes' answers), and game theory (counting wins in repeated games).
The binomial model assumes: (1) A fixed number of trials n, (2) Each trial is independent of the others, (3) Each trial has exactly two outcomes (success or failure), and (4) The probability of success p is constant for every trial. If these assumptions are violated, the binomial model may not be appropriate.
P(X = k) is the probability of getting exactly k successes—no more, no less. P(X ≤ k) is the cumulative probability of getting k or fewer successes, which includes all outcomes from 0 through k. The cumulative probability is always greater than or equal to the exact probability for the same k.
Use P(X ≥ k) when you want to know the probability of getting at least k successes. This is common in scenarios like 'What's the chance of passing if I need at least 7 correct answers?' P(X ≤ k) is for 'at most' questions, while P(X ≥ k) is for 'at least' questions.
The mean μ = n × p represents the expected number of successes if you were to repeat the experiment many times. For example, if you flip a fair coin 100 times (n=100, p=0.5), you expect about 50 heads on average. Individual experiments will vary around this expected value.
The standard deviation σ = √(n × p × (1-p)) measures the typical spread of the number of successes around the mean. About 68% of outcomes fall within one standard deviation of the mean, and about 95% fall within two standard deviations. A larger standard deviation means more variability in your results.
For very large n, the binomial distribution approaches the normal distribution (by the Central Limit Theorem). When n is extremely large, computing exact binomial probabilities can be slow and may have numerical precision issues. For n > 200, consider using the normal approximation instead.
When p = 0, every trial fails, so P(X = 0) = 1 and all other probabilities are 0. When p = 1, every trial succeeds, so P(X = n) = 1 and all other probabilities are 0. These are degenerate cases where there's no randomness—the outcome is certain.
This calculator is an educational tool designed to help you understand binomial distributions and verify your work. While it provides accurate calculations, you should use it to learn the concepts and check your manual calculations, not as a substitute for understanding the material. Always verify important results independently.
The calculator uses standard floating-point arithmetic with results displayed to 6 decimal places. For most practical purposes, this precision is more than sufficient. Very small probabilities (less than 10^-10) may be displayed as 0 due to numerical limitations.
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Enter the number of trials (n), probability of success (p), and select a probability query to explore the binomial distribution.