Read it again any time, including while you work through the exercises.
You repeat the same two-outcome experiment (success or failure) 4 times, identically and independently: this is a Bernoulli scheme. You already know a tree lets you compute probabilities. But with 4 repetitions, the tree already has 16 paths... and with 10 repetitions, 1024! We need a shortcut to count paths without drawing them.
That shortcut is the binomial coefficient, written C(n, k) and read 'n choose k'. It counts the number of paths in the tree that contain exactly k successes among n repetitions. For example, C(4, 2) = 6: there are 6 ways to place 2 successes in a sequence of 4 trials (SSFF, SFSF, SFFS, FSSF, FSFS, FFSS).
To find these coefficients without a fancy formula, use Pascal's triangle: each row starts and ends with 1, and each inner number is the sum of the two numbers just above it. Row n = 4 reads: 1, 4, 6, 4, 1. Notice the symmetry: C(n, k) = C(n, n - k), because choosing the k successes amounts to choosing the n - k failures.
Everything comes together in the binomial distribution formula. If X counts the number of successes in a Bernoulli scheme with parameters n and p, then P(X = k) = C(n, k) × p^k × (1 - p)^(n - k). The logic: each path with k successes has probability p^k × (1 - p)^(n - k), and there are C(n, k) such paths.
Checking reflex: a binomial coefficient is a WHOLE number (we are counting paths), and a probability always stays between 0 and 1. The most frequent slip? Forgetting the coefficient: without it, you count only one path instead of all of them.