Counting successes

Start with the simplest possible random variable: something happens, or it doesn’t.

Call the outcome you are counting a success, with probability p, and everything else a failure, with probability 1 − p. Let X be 1 for a success and 0 for a failure. That is a Bernoulli random variable, and its whole distribution fits in two lines.

Its mean and variance come straight out of last lesson’s formulas, and both are worth memorising because the binomial builds on them.

Note that E(X²) equals p as well, because 0² = 0 and 1² = 1 leave the values unchanged. That is what makes the variance collapse so neatly.

“Success” carries no judgement: if you are counting faulty items, a faulty item is the success. Deciding what counts as a success, and getting p to match it, is the first move in every question in this lesson.

Now repeat the trial n times, independently, with p unchanged each time, and count the successes.

Take 5 free throws with p = 0.6 and ask for the probability of exactly 3 successes. One way it could happen is success, success, success, fail, fail, whose probability is 0.6 × 0.6 × 0.6 × 0.4 × 0.4, that is 0.6³ × 0.4². But so does fail, success, fail, success, success, with probability 0.4 × 0.6 × 0.4 × 0.6 × 0.6, which multiplies out to exactly the same number.

Every arrangement with 3 successes has the same probability. So the total is that probability multiplied by the number of arrangements, and counting arrangements is what the combination \(\binom{n}{x}\) does.

Read the formula as three parts: \(p^x\) for the successes, \((1-p)^{n-x}\) for the failures, and \(\binom{n}{x}\) for the number of ways to arrange them. Written compactly, this is \(X\sim\text{Bin}(n,p)\), and stating n and p explicitly is the first line of any solution.

Three conditions must hold before you may use it: a fixed number of trials, exactly two outcomes each time, and independence with p constant throughout. Drawing cards without replacement fails the last condition, because p changes as the deck depletes.

Change n and p and watch what happens to the distribution.

Three things to notice. With p = 0.5 the distribution is symmetric. Push p towards 0 or 1 and it becomes lopsided, bunching against the end. And as n grows the shape becomes smoother and more bell-like, which is the hint that connects this topic to the normal distribution next term.

The centre and spread have short formulas, which follow from adding up n independent Bernoulli variables:

The mean is exactly what intuition says: 8 free throws at 60% should produce about 4.8 goals. The variance being np(1 − p) explains something less obvious, that uncertainty is greatest when p is near 0.5 and shrinks as p approaches either extreme. A 99% reliable machine is predictable; a coin flip is not.

Next: the working method, twenty practice questions, ten harder ones, and the ClassPad guide covering binomialPDf, binomialCDf and finding n.