What makes a question binomial
Four things have to be true, and it is worth checking all four rather than three:
- A fixed number of tries. You decided on ten before you started, not "however many it takes".
- Two outcomes. It worked or it did not.
- The same chance every time. Not warming up, not getting tired.
- Independence. One attempt does not change the next.
The last two are where real situations break the model, and usually for the same reason: life has memory and the formula does not. A shooter who is tired by the ninth shot is not binomial. Neither is a machine that jams more often once it has jammed. That does not make the answer useless, but it does make it a starting point rather than a verdict.
The formula
Read left to right, it says: pick which k of your n tries were the successful ones, multiply by the chance those all worked, then by the chance the rest all failed. The C(n, k) at the front is doing the counting, and it is the reason ten successes out of twenty is far more likely than twenty out of twenty.
Worked example
A basketball player who makes 80 percent of his free throws takes ten.
The most likely single outcome is exactly eight made, at 30.2%. But that is not what people notice. What they notice is the misses:
P(making at least 9 of 10) = 0.2684 + 0.1074 = 37.6%
Which means that 62.4% of the time, an 80 percent shooter misses two or more. Missing twice in ten is not a slump. It is the majority outcome, and it is what an eighty percent shooter looks like on a completely ordinary night.
The most useful thing this distribution does
People are reliably wrong about at least one, and always in the same direction.
Ten tries at a 10 percent chance each feels like it should be a certainty. It is not. The chance of failing all ten is 0.910 = 0.349, so the chance of at least one success is about 65 percent. Roughly one time in three you get nothing at all.
| Tries at 10% each | Chance of at least one success |
|---|---|
| 5 | 41.0% |
| 10 | 65.1% |
| 20 | 87.8% |
| 50 | 99.5% |
The trick worth keeping is that at least one is easiest to reach by calculating none and subtracting. That is one term instead of many, and it is how you do it in your head.
Binomial or hypergeometric
The two answer nearly the same question and differ on one point: whether you put things back.
Binomial assumes every trial is identical, which is true for coins and dice and free throws. Hypergeometric assumes a finite pool that shrinks as you draw, which is true for cards, shipments and audit samples. They share the same mean exactly, and the hypergeometric one is always the less variable.
The practical rule: if you are drawing from a pool and your sample is more than about 5 percent of it, use hypergeometric. Below that the two converge and either will do.
When a normal curve is close enough
For large n the binomial starts to look like a normal distribution, which is why so many textbooks reach for that approximation. The usual condition is that both np and n(1 − p) are at least 5.
It is a decent rule. At n = 20 and p = 0.5 the approximation lands within half a percent; at n = 200 it is within a twentieth of a percent. But it is worth being clear that the approximation exists to save arithmetic that nobody has to do by hand any more. This page computes the exact answer at any size, so treat the normal curve as a way of understanding the shape rather than as something you need.
If you do want the normal distribution itself, the normal distribution calculator handles it directly, and the combination calculator covers the C(n, k) term on its own.