Probability Calculator

Most probability questions you encounter — independent events, conditional probability, at-least-one calculations — have simple formulas, but applying them to a specific problem is where most students get stuck.

The calculator handles the common cases: combining independent events (multiply), mutually exclusive events (add), conditional probability (Bayes' theorem), at-least-one (1 minus probability of none), and binomial outcomes (n trials, k successes). Each mode has its own input form and shows the formula being used so it's a teaching tool, not a black box.

The trickiest type to get right is conditional — P(A given B) is not the same as P(B given A), and confusing them is the error behind a lot of bad reasoning in news articles about medical tests and forensic evidence. The Bayes mode is set up to make the distinction visible.

The core formulas are short. Independent events: P(A and B) = P(A) × P(B). Mutually exclusive: P(A or B) = P(A) + P(B). General union: P(A or B) = P(A) + P(B) − P(A and B). Conditional: P(A given B) = P(A and B) / P(B). Bayes: P(A given B) = P(B given A) × P(A) / P(B). Binomial: P(k successes in n trials) = C(n,k) × p^k × (1-p)^(n-k), where C(n,k) is the binomial coefficient.

The pain this addresses: probability questions in real life are confusing because they mix conditional probabilities people misread. The classic example: a disease affects 1 in 1000 people. The test is 99% accurate. You test positive. What's the chance you have the disease? Intuition says ~99%. The actual answer (Bayes' theorem) is about 9%. Most positives are false positives because the underlying rate is so low. The calculator runs this kind of math without you having to remember the formula, which prevents the dangerous miscalibration that comes from doing it wrong in your head.

Worked example: rolling four six-sided dice, what's the probability of at least one 6? P(no 6 on one die) = 5/6. P(no 6s on four dice) = (5/6)^4 = 0.482. P(at least one 6) = 1 − 0.482 = 0.518. Just over 50%. This is the foundation of Chevalier de Méré's gambling problem from the 1600s — same calculation, just for showing up at least once across N attempts. The 'at least one' framing trips students up; the 'subtract from 1' approach is much faster than summing exact-1 + exact-2 + exact-3 + exact-4.

Where this can go subtly wrong: assuming independence. Card draws without replacement aren't independent — drawing an ace lowers the probability of drawing another. Two events from the same population usually aren't independent. The calculator multiplies for independent events; if your real-world events have any correlation, the result is wrong in the direction of overconfidence (correlated risks happen together more often than the math suggests). When in doubt, model the dependence explicitly or use simulation.