Binomial distribution: the formula you need for exams

The formula

If $X$ follows a binomial distribution $\mathcal{B}(n, p)$, then:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

• $n$ = number of trials (repeated)
• $p$ = probability of success on each trial
• $k$ = number of successes wanted
• $\binom{n}{k}$ = binomial coefficient (ways to choose $k$ successes among $n$)

The 3 conditions (always check)

1. Repetition: the trial is repeated $n$ times.
2. Independence: each trial is independent of the others.
3. Two outcomes only: success (prob. $p$) or failure (prob. $1-p$).

If any is missing → NOT binomial (often hypergeometric if drawing without replacement).

Mean and variance

$E(X) = np \quad ; \quad V(X) = np(1-p)$

Useful sanity check.

Step-by-step example

Fair die rolled 5 times. Probability of getting exactly 2 sixes?

• $n = 5$, $p = 1/6$, $k = 2$.
• Conditions: yes (independent rolls, 2 outcomes six / not-six).
• $P(X = 2) = \binom{5}{2} (1/6)^2 (5/6)^3$
• $\binom{5}{2} = 10$
• $= 10 \times \dfrac{1}{36} \times \dfrac{125}{216} = \dfrac{1250}{7776} \approx 0.161$

About 16.1%.

On a calculator

• TI / Casio: distributions menu, choose binompdf (for $P(X = k)$) or binomcdf (for $P(X \leq k)$).
• Arguments: $n$, $p$, $k$.

Common exam cases

• $P(X \geq k) = 1 - P(X \leq k-1)$ (complement)
• $P(X \leq k) =$ sum from 0 to $k$ (or use binomcdf directly)

Common mistake

Confusing $p$ (success) with $q = 1-p$ (failure). Always clearly define what counts as a success.

?À toi de jouer

A fair coin is flipped 10 times. Probability of exactly 3 heads?

A.≈ 0.117 (11.7%)B.0.3C.≈ 0.375D.≈ 0.001

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