Repeat the same trial $n$ times, independently, each succeeding with the same chance $p$, and count the successes. That is a binomial experiment, and it is the most common distribution on any exam paper. Its centre comes first: $\mu = np$, read mu equals n p, where $\mu$ — the Greek letter mu, said “mew” — is the expected number of successes, $n$ is the count of trials, and $p$ is the per-trial chance. Twenty guesses at a four-choice question centre on $20 \times 0.25 = 5$ right answers. Compute that first, always; it is the rail every later answer gets checked against.

For an exact count, $P(X = k) = \binom{n}{k} p^{k} (1-p)^{\,n-k}$, read P of X equals k equals n choose k, times p to the k, times one minus p to the n minus k. Read it in three pieces. $\binom{n}{k}$ — that is ${}_{n}C_{k}$ in its stacked clothing — counts WHICH trials succeeded. $p^{k}$ prices those $k$ successes. $(1-p)^{\,n-k}$ prices the $n-k$ failures, because the others must MISS, and missing is not free. Drop the $\binom{n}{k}$ and you have priced exactly one pattern out of many — the single most common error in this lesson.

Spread comes last: $\sigma^{2} = np(1-p)$, the variance of the success count, where $\sigma$ (sigma) is the standard deviation and $\sigma^{2}$ its square. Note where $p(1-p)$ peaks — at $p = 0.5$. A fair coin is the most unpredictable coin there is, and a near-certain trial barely wobbles at all.