Bernoulli and binomial distributions

The binomial distribution counts successes in a fixed number n of independent trials that each succeed with the same probability p.

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \quad k = 0, 1, \dots, n

Binomial pmf (Bernoulli is the case n = 1)

$n$: number of trials (fixed in advance)
$p$: probability of success on each trial
$k$: number of successes
$\binom{n}{k}$: number of ways to choose which $k$ trials succeed

\E(X) = np, \qquad \Var(X) = np(1-p), \qquad \SD(X) = \sqrt{np(1-p)}

Mean and variance

$X$: number of successes, $X \sim \text{Binomial}(n, p)$

When to use

Counting successes (conversions, defective items, correct guesses) out of a known number of trials. A single trial (success = 1, failure = 0) is a Bernoulli variable with mean $p$ and variance $p(1-p)$ .

Assumptions

The "BINS" conditions: Binary outcome on each trial; Independent trials; Number of trials fixed; Same probability $p$ on every trial. Sampling without replacement from a small population breaks independence; use the hypergeometric instead (the binomial is a good approximation when the sample is under about 10% of the population).

How to interpret

The distribution is symmetric when $p = 0.5$ and skewed towards the longer tail otherwise.
The proportion of successes $X/n$ has mean $p$ and SD $\sqrt{p(1-p)/n}$ .
Normal approximation: reasonable when $np \ge 10$ and $n(1-p) \ge 10$ ; the engine computes exact binomial probabilities, so the approximation is only needed for intuition.

Common mistakes

"At least 3" is $P(X \ge 3) = \text{sf}(2)$ ; the engine's sf(k) is $P(X > k)$ .
Using the binomial for a count with no fixed upper limit (use Poisson) or for "trials until the first success" (use geometric).
Forgetting the $\binom{n}{k}$ factor: $p^k(1-p)^{n-k}$ is the probability of one particular sequence.

Example

$n = 10$ calls, each converting with $p = 0.2$ . $P(X = 3) = 0.2013$ ; $P(X \le 2) = 0.6778$ ; $P(X \ge 3) = \text{sf}(2) = 0.3222$ . Mean $= 2$ , SD $= \sqrt{1.6} \approx 1.265$ .