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High school🎲Probability12 min read
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Binomial distribution: complete lesson (Bernoulli, formula, expectation)

n independent 2-outcome trials. P(X = k), expectation np, variance np(1βˆ’p). Full grade-12 lesson.

Prerequisites :

  • Conditional probability
  • Binomial coefficient

Bernoulli trial

A Bernoulli trial is a random experiment with 2 outcomes:

  • β€œSuccess” with probability pp.
  • β€œFailure” with probability q=1βˆ’pq = 1 - p.

Examples: coin flip, drawing a card (red or black), testing if an item works.

Bernoulli scheme

nn Bernoulli trials:

  • Identical (same pp).
  • Independent.

Let XX be the number of successes. Then XX follows a binomial distribution with parameters nn and pp: X∼B(n,p)X \sim \mathcal{B}(n, p).

Binomial formula

P(X=k)=(nk)pk(1βˆ’p)nβˆ’kP(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}

for k=0,1,2,…,nk = 0, 1, 2, \ldots, n.

Interpretation:

  • (nk)\binom{n}{k}: number of ways to get kk successes among nn slots.
  • pkp^k: probability of kk successes.
  • (1βˆ’p)nβˆ’k(1-p)^{n-k}: probability of nβˆ’kn-k failures.

Expectation and variance

E(X)=nβ‹…pE(X) = n \cdot p

V(X)=nβ‹…pβ‹…(1βˆ’p)V(X) = n \cdot p \cdot (1 - p)

Οƒ(X)=np(1βˆ’p)\sigma(X) = \sqrt{n p (1 - p)}

Meaning: over nn trials with success probability pp, expect npnp successes.

Typical example

Flip a fair coin 10 times. Probability of exactly 3 heads?

X∼B(10,0.5)X \sim \mathcal{B}(10, 0.5), k=3k = 3.

P(X=3)=(103)Γ—0.53Γ—0.57=120Γ—0.510β‰ˆ0.117P(X = 3) = \binom{10}{3} \times 0.5^3 \times 0.5^7 = 120 \times 0.5^{10} \approx 0.117

Cumulative: P(X≀k)P(X \leq k)

For an inequality, sum up:

P(X≀k)=P(X=0)+P(X=1)+…+P(X=k)P(X \leq k) = P(X = 0) + P(X = 1) + \ldots + P(X = k)

For P(Xβ‰₯k)=1βˆ’P(X≀kβˆ’1)P(X \geq k) = 1 - P(X \leq k - 1) (complement).

Graph

Bar chart: heights = P(X=k)P(X = k). Symmetric if p=0.5p = 0.5, skewed otherwise.

Special cases

  • n=1n = 1: XX is a Bernoulli variable (X∈{0,1}X \in \{0, 1\}). E(X)=pE(X) = p, V(X)=p(1βˆ’p)V(X) = p(1-p).
  • p=0p = 0: X=0X = 0 always.
  • p=1p = 1: X=nX = n always.

Application: normal approximation

For large nn with pp not too close to 00 or 11, the binomial approaches a normal distribution with mean npnp and standard deviation np(1βˆ’p)\sqrt{np(1-p)}.

Common mistakes

  1. Forgetting independence of trials.
  2. Writing E(X)=pE(X) = p instead of npnp.
  3. Confusing (nk)\binom{n}{k} with nkn^k in the formula.
  4. Using binomial when trials aren’t identical (e.g., drawing without replacement = hypergeometric).
#binomial distribution#Bernoulli#expectation#variance#grade 12#lesson