Binomial Probability Calculator

What makes a probability problem binomial

A binomial situation has four specific features, worth checking before reaching for this formula. There's a fixed number of trials decided in advance, not "keep flipping until heads." Each trial has only two outcomes that matter: success or failure. The probability of success stays the same on every trial. And the trials don't affect each other, so trial 3 doesn't care what happened on trial 2.

Drawing cards without putting them back breaks the third and fourth conditions, since the deck composition changes each draw. That's a hypergeometric problem, not a binomial one, even though it looks similar on the surface.

The binomial formula, piece by piece

C(n,k), read as "n choose k," is doing more work than it looks like. Flip a coin 5 times and want exactly 3 heads: there isn't just one way that happens. HHHTT, HHTHT, HTHHT, and several others all count. C(5,3) = 10 tells you there are 10 different orderings that give exactly 3 heads, and the formula needs to count all of them, not just one.

For "at least k" or "at most k," you're adding the exact-k formula across a range of values rather than running it once. At least 2 successes out of 10 trials means adding P(X=2) + P(X=3) all the way up to P(X=10). The calculator above does this sum automatically and shows every individual term in the distribution chart.

A worked example by hand

What's the probability of getting exactly 3 heads in 5 coin flips? n = 5, k = 3, p = 0.5.

First, C(5,3) = 5! / (3! × 2!) = 120 / (6 × 2) = 10. Then p^k = 0.5³ = 0.125, and (1−p)^n−k = 0.5² = 0.25.

Multiply all three together: 10 × 0.125 × 0.25 = 0.3125, or 31.25%.

Notice this is higher than the roughly 6.25% chance of any single specific sequence like HHHTT. That's the C(n,k) term doing its job, accounting for all 10 orderings that count as "3 heads" rather than just one.

Where the mean and variance come from

A binomial distribution has a predictable average and spread, useful for sanity-checking a result. The expected number of successes is n × p. Flip a coin 100 times and you'd expect 50 heads on average, even though any individual run lands above or below that. The variance is n × p × (1−p), which measures how much individual runs typically deviate from that average.

Common mistakes with binomial probability

Forgetting the C(n,k) term entirely. Multiplying just p^k × (1−p)^n−k gives the probability of one specific ordering, not the probability of getting k successes in any order, which is almost always what the question actually asks.

Using it when trials aren't independent or p changes. Card draws without replacement, or scenarios where earlier results change later odds, don't qualify. The formula still returns a number, it just won't mean anything.

Mixing up "at least" and "more than." "At least 2" includes exactly 2. "More than 2" starts at 3. This single-word difference changes which terms get summed and is a frequent source of wrong answers.

Questions people actually ask about this