The binomial distribution

Many questions look different and are really the same: how likely are exactly 2 sixes in 10 rolls? How likely are exactly 3 faulty parts among 20? How likely is it to score 5 out of 8 free throws? Every time the same trial is repeated, each time with the same chance, and all that counts is how often it worked.

The Bernoulli chain

Three conditions have to hold. First: every trial has only two outcomes, hit or no hit. Second: the hit probability $p$ stays the same in every trial. Third: the trials do not influence each other. If all of that holds, you have a Bernoulli chain, and then the formula of this topic applies.

Where the formula comes from

Take 3 rolls with $p = \frac{1}{6}$ and ask for exactly one six. One possible course is hit, miss, miss, and its probability is $\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{5}{6}$. But the six could also come on the second or third roll. All three courses have the same probability, so you have to multiply by 3.

That is exactly what the binomial coefficient $\binom{n}{k}$ does, read as $n$ choose $k$. It counts in how many ways $k$ hits can be spread across $n$ trials. For 3 trials and 1 hit that is 3 ways, for 10 trials and 2 hits already 45.

The three building blocks are easy to remember. $p^k$ stands for the $k$ hits, $(1-p)^{n-k}$ for the remaining trials without a hit, and the binomial coefficient for the fact that the hits can sit in different places. For 10 rolls and exactly 2 sixes that gives $\binom{10}{2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^8 \approx 0.29$.

At least and at most

It is rare for a question to ask about exactly one number. For at least one hit the direct calculation would be tedious, because you would have to add every case from 1 to $n$. The detour via the complementary event is far shorter: at least one hit is the opposite of no hit at all, so $P(X \geq 1) = 1 - P(X = 0)$. And $P(X = 0)$ is simply $(1-p)^n$.