Solving binomial probability comes down to one formula, applied once you’ve confirmed your problem meets four specific conditions. Whether you need the exact probability of a single outcome or a cumulative “at least” or “at most” answer, the process follows the same logic. Here’s how to work through it from start to finish.
Confirm Your Problem Is Binomial
Before plugging anything into a formula, check that your scenario meets all four requirements of a binomial distribution. If any one of these is missing, the binomial formula won’t give you a valid answer.
- Fixed number of trials (n): You know in advance how many times the experiment happens. Ten coin flips, 15 free throws, 50 inspected parts. The number doesn’t change midway through.
- Two possible outcomes per trial: Each trial results in either a “success” or a “failure.” These labels are flexible. A success could be heads, a made free throw, a defective part, or anything you’re counting.
- Constant probability of success (p): The chance of success stays the same from one trial to the next. If you’re drawing from a deck of cards without replacing them, the probability shifts each draw, so that wouldn’t qualify.
- Independent trials: The outcome of one trial doesn’t influence the next. A coin doesn’t “remember” it landed heads last time.
If all four conditions hold, you’re working with a binomial distribution and can move on to the formula.
The Binomial Probability Formula
The formula calculates the probability that a success occurs exactly k times out of n trials:
P(X = k) = [n! / (k!(n − k)!)] × p^k × (1 − p)^(n − k)
Here’s what each piece means:
- n = the total number of trials
- k = the number of successes you’re solving for
- p = the probability of success on a single trial
- (1 − p) = the probability of failure on a single trial
The front portion, n! / (k!(n − k)!), is called “n choose k.” It counts the number of different ways k successes can be arranged across n trials. The exclamation mark means factorial: 5! = 5 × 4 × 3 × 2 × 1 = 120. The rest of the formula multiplies the probability of getting exactly k successes by the probability of getting the remaining (n − k) failures.
A Worked Example
Suppose a city solves 20% of its petty crime cases. Three new cases come in. What’s the probability that exactly two of the three are solved?
Identify your variables first: n = 3 (three cases), k = 2 (two solved), p = 0.20 (20% solve rate), and (1 − p) = 0.80.
Start with the combination. “3 choose 2” = 3! / (2! × 1!) = 6 / (2 × 1) = 3. This tells you there are three different ways two successes can occur among three trials.
Now multiply: 3 × (0.20)^2 × (0.80)^1 = 3 × 0.04 × 0.80 = 0.096. The probability of exactly two cases being solved is 9.6%.
Solving “At Most” and “At Least” Problems
Many real problems don’t ask for an exact number of successes. They ask things like “at most 4” or “at least 3.” These require adding up multiple individual probabilities.
“At Most” Problems
“At most k” means k or fewer successes. You sum the probabilities from 0 up through k. For example, “at most 2 successes out of 5 trials” means you calculate P(X = 0) + P(X = 1) + P(X = 2) and add them together.
“At Least” Problems
“At least k” means k or more successes. You could sum every probability from k up to n, but there’s a faster approach: use the complement. “At least k” is the opposite of “at most k − 1,” so:
P(X ≥ k) = 1 − P(X ≤ k − 1)
If you need the probability of at least 3 successes out of 10 trials, calculate the cumulative probability of 2 or fewer successes, then subtract that from 1. This saves you from computing eight separate terms.
Using a Graphing Calculator
Working the formula by hand gets tedious once n climbs past single digits. Graphing calculators like the TI-84 have two built-in functions that handle the arithmetic instantly.
binompdf(n, p, k) gives you the probability of exactly k successes. The “pdf” stands for probability density function. You enter the number of trials, the probability of success, and the specific number of successes you want. For example, binompdf(7, 0.35, 4) returns the probability of exactly 4 successes in 7 trials with a 35% success rate per trial.
binomcdf(n, p, k) gives you the cumulative probability of k or fewer successes. It adds up every individual probability from 0 through k in one step. For “at most” problems, you use binomcdf directly. For “at least” problems, subtract from 1: enter 1 − binomcdf(n, p, k − 1). Notice you adjust k down by one, because “at least 4” is the complement of “at most 3.”
Using Excel or Google Sheets
Spreadsheet software uses a single function with a toggle for exact vs. cumulative calculations:
BINOM.DIST(number_s, trials, probability_s, cumulative)
The four arguments are: the number of successes, the number of trials, the probability of success per trial, and either TRUE or FALSE. Set the last argument to FALSE to get the probability of that exact number of successes (like binompdf). Set it to TRUE to get the cumulative probability of that many successes or fewer (like binomcdf). For “at least” problems in Excel, use the same complement trick: =1 − BINOM.DIST(k − 1, n, p, TRUE).
Finding the Mean and Standard Deviation
Sometimes you need to describe the overall shape of a binomial distribution rather than solve for a single probability. Two quick formulas handle this:
- Mean (expected value): n × p
- Variance: n × p × (1 − p)
The standard deviation is the square root of the variance. If you flip a fair coin 100 times, the expected number of heads is 100 × 0.50 = 50, the variance is 100 × 0.50 × 0.50 = 25, and the standard deviation is 5. This tells you that results within roughly 5 heads of 50 are common, while results far outside that range are unusual.
When to Use the Normal Approximation
For large numbers of trials, calculating binomial probabilities term by term becomes impractical even with a calculator. You can approximate the result using a normal (bell curve) distribution when both n × p and n × (1 − p) are at least 5. Check the smaller of the two values first; if it hits 5, the larger one will too.
To use this shortcut, treat the binomial distribution as a normal distribution with the same mean (n × p) and standard deviation (square root of n × p × (1 − p)). Then convert your value of k to a z-score using z = (k − mean) / standard deviation, and look up the probability on a standard normal table or use a calculator’s normalcdf function. Add a continuity correction of 0.5 when converting from the discrete binomial to the continuous normal curve: if you want P(X ≤ 8), use 8.5 in your z-score calculation, and if you want P(X ≥ 8), use 7.5.
Common Mistakes to Avoid
The most frequent error is misidentifying what counts as a “success.” The term doesn’t have to mean something good. If you’re calculating the probability of defective parts, a defective part is your success. Define it clearly before you start.
Another common mistake is confusing “at least” with “more than.” “At least 3” includes 3 itself (≥ 3), while “more than 3” does not (> 3, meaning 4 or higher). This one-unit difference changes your complement calculation and your answer.
Finally, watch your factorials. By definition, 0! = 1, not 0. Forgetting this will break your combination calculation any time k equals 0 or n, which are perfectly valid scenarios. The probability of zero successes in n trials is simply (1 − p)^n, and the combination term equals 1.