Expectation of binomial: E(X) = np, V(X) = np(1−p)

The 3 formulas to remember

For $X \sim \mathcal{B}(n, p)$:

$E(X) = n p$

$V(X) = n p (1 - p)$

$\sigma(X) = \sqrt{n p (1 - p)}$

The intuition

• Over $n$ trials with $p$ success probability, expect $np$ successes on average.
• Spread is maximal when $p = 0.5$ (most uncertain results).

Example: 100 coin flips ($p = 0.5$).

• $E(X) = 100 \times 0.5 = 50$ heads expected.
• $\sigma(X) = \sqrt{100 \times 0.25} = 5$ (variability of 5 heads around 50).

Sanity checks

• 1 coin flip: $E = 1 \times 0.5 = 0.5$. Correct (average between 0 and 1).
• Rolling “6” on 6 dice throws: $p = 1/6$, $E = 6 \times 1/6 = 1$. Expect 1 six.

Practical use

Quality tester: on 1000 items with 2 % defects ($p = 0.02$), expect $E = 1000 \times 0.02 = 20$ defects. Std dev $\sigma = \sqrt{1000 \times 0.02 \times 0.98} \approx 4.4$.

Why these formulas?

Because a $\mathcal{B}(n, p)$ is a sum of $n$ independent Bernoulli variables with parameter $p$:

• Bernoulli: $E = p$, $V = p(1-p)$.
• Sum of $n$: $E = np$, $V = np(1-p)$ (variance is additive for independent).

Common mistakes

1. Writing $E(X) = p$ (missing $n$).
2. Writing $V(X) = np$ (missing $(1-p)$).
3. Confusing std dev and variance (mind the $\sqrt{}$).
4. Using it for without-replacement (that’s hypergeometric).

?À toi de jouer

X ~ B(50, 0.3). E(X) = ?

A.15B.10.5C.50D.0.3

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