Counting successes in a fixed number of trials
The binomial answers: n independent trials, each succeeding with probability p: what is the chance of exactly k successes, at most k, at least k? The formula has three moving parts: C(n, k) counts which trials succeed, pk prices the successes, (1−p)n−k prices the failures.
Ten fair coin flips. Exactly 3 heads: C(10, 3) × 0.510 = 0.1172. Three or fewer: 0.1719. Add the bars for 0, 1, 2, 3. Seven or more, by symmetry, also 0.1719. The chart on this page is the whole distribution; every binomial question is "which bars do I add?", exactly/at most/at least/between are just four ways of choosing bars.
The four conditions (BINS)
- Binary outcomes: success or failure, nothing between.
- Independent trials: one result does not sway the next.
- N fixed in advance: you decided the number of trials before starting.
- Same p on every trial: the coin does not change.
Break the fixed-n condition (flipping until the third head arrives) and you have left binomial country: that distribution is the negative binomial.
"At least 1" computed the long way. Use the complement: 1 − P(0). One subtraction beats summing n bars, on this page and everywhere else in probability.
Frequently asked questions
Does 'success' have to mean something good?
No, success is just the outcome you are counting. Defects, rainy days and missed buses are all binomial successes if you define them so.
When is the normal approximation acceptable?
When np and n(1−p) both reach about 10, the binomial bell is nearly normal: that is what the chart shows as n grows.