Count the paths, and a shape appears.

Flip a coin three times and there are eight equally likely stories. Sort them by how many heads they contain and the binomial distribution falls out — no formula needed yet, just counting.

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Every route through the tree has probability $\left(\tfrac12\right)^3 = \tfrac18$. One path gives three heads, three give two, three give one, one gives none — the row $1,\ 3,\ 3,\ 1$ from Pascal's triangle in Chapter 7 of Pure.

Gather the paths by their number of heads and the counts become heights. That is $X \sim B(3,\,0.5)$: the coefficient counts the orders, and each order carries $\tfrac18$.

With 20 tosses there are over a million paths, but the same counting holds, and the shape settles into a hump around ten heads.

Nothing about the argument needed a fair coin. Slide $p$ down to 0.3 and the $p^r(1-p)^{n-r}$ does the rest: the hump slides left and leans, because low counts are now the likely ones.

Four conditions, one formula, a calculator for the rest.

Discrete random variables, the binomial conditions, cumulative probabilities, judging whether the model fits, and twelve graded exam questions with full worked solutions are waiting.

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