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What is the binomial distribution, and what are its mean and variance?

Define the binomial distribution B(n,p)B(n, p), state its probability mass function, and find its mean npn p and variance np(1−p)n p (1 - p)

A focused answer to the HSC Maths Extension 1 dot point on the binomial distribution. The pmf P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}, mean npn p, variance np(1−p)n p (1 - p), and standard situations that fit the model.

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What this dot point is asking

NESA wants you to recognise scenarios that fit the binomial model (fixed number of independent Bernoulli trials with the same success probability), state the probability mass function, and compute the mean, variance and standard deviation.

The answer

Definition

Let XX be the number of successes in nn independent Bernoulli trials, each with success probability pp. Then XX has the binomial distribution B(n,p)B(n, p) (or Bin(n,p)\text{Bin}(n, p)).

The parameters are:

  • nn: the number of trials (a positive integer).
  • pp: the success probability on each trial (0≤p≤10 \le p \le 1).
  • q=1−pq = 1 - p: the failure probability.

The probability mass function

For k=0,1,2,…,nk = 0, 1, 2, \dots, n,

P(X=k)=(nk)pk(1−p)n−k.P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}.

The binomial coefficient (nk)\binom{n}{k} counts the number of ways to arrange kk successes among nn trials.

Mean and variance

By summing nn independent Bernoulli trials (each with mean pp and variance p(1−p)p(1 - p)):

E(X)=np,Var(X)=np(1−p),σ=np(1−p).E(X) = n p, \qquad \text{Var}(X) = n p (1 - p), \qquad \sigma = \sqrt{n p (1 - p)}.

Conditions for the binomial model

Four conditions must hold:

  1. A fixed number of trials, nn.
  2. Each trial has exactly two outcomes (success or failure).
  3. The trials are independent.
  4. The probability of success is the same for every trial.

If any of these fails, the distribution is not binomial.

Common scenarios

  • Number of heads in nn coin flips: X∼B(n,p)X \sim B(n, p) with pp the heads probability.
  • Number of items defective in a batch of nn (with replacement, or with a large enough population to treat as approximately independent).
  • Number of correct answers on a multiple-choice test if every question is guessed.
  • Number of successful free throws in a fixed number of attempts (independence is a simplifying assumption).

Symmetric and skewed cases

  • If p=12p = \tfrac{1}{2}, the distribution is symmetric around n2\frac{n}{2}.
  • If p<12p < \tfrac{1}{2}, the distribution is right-skewed (long tail to the right).
  • If p>12p > \tfrac{1}{2}, it is left-skewed.

Binomial probability mass function for three values of p Three small bar charts of P X equals k against k for n equals ten trials. p equals zero point two is right skewed. p equals zero point five is symmetric. p equals zero point eight is left skewed. p = 0.2 p = 0.5 p = 0.8 Mean shifts with p; symmetric when p = 0.5; skewed otherwise.

The skew decreases as nn grows; for large nn, the binomial approaches the normal (see the normal approximation dot point).

Building one distribution, stage by stage

To see where the pmf, the mean and the variance come from, take a concrete case: a biased coin with p=0.4p = 0.4 flipped n=10n = 10 times, so X∼B(10,0.4)X \sim B(10, 0.4). The three panels build the whole distribution from the formula and then read its centre and spread off the picture.

Stage 1, compute a few probabilities from the formula. Each bar height is P(X=k)=(10k)(0.4)k(0.6)10−kP(X = k) = \binom{10}{k}(0.4)^k(0.6)^{10 - k}. For the first few values, P(X=0)=(0.6)10≈0.006P(X = 0) = (0.6)^{10} \approx 0.006, P(X=1)=(101)(0.4)(0.6)9≈0.040P(X = 1) = \binom{10}{1}(0.4)(0.6)^9 \approx 0.040, and P(X=2)=(102)(0.4)2(0.6)8≈0.121P(X = 2) = \binom{10}{2}(0.4)^2(0.6)^8 \approx 0.121. Each is one number from the same formula.

Binomial distribution B(10, 0.4)A bar chart of P(X equals k) for a binomial distribution with ten trials and success probability 0.4. The tallest bars are at k equals 4, near the mean of 4.kP(X = k)00.00610.04020.121345678910Stage 1Each bar height is P(X = k) from the formula nCr p^k q^(n-k); here k = 0, 1, 2.

Stage 2, fill in every kk to get the whole distribution. Repeating the formula for k=0,1,…,10k = 0, 1, \dots, 10 gives all eleven bars. Because the eleven probabilities are the terms of (0.6+0.4)10(0.6 + 0.4)^{10}, they sum to 110=11^{10} = 1: the distribution is complete and valid.

Binomial distribution B(10, 0.4)A bar chart of P(X equals k) for a binomial distribution with ten trials and success probability 0.4. The tallest bars are at k equals 4, near the mean of 4.kP(X = k)012345678910Stage 2Filling in every k from 0 to 10 gives the whole distribution; heights sum to 1.

Stage 3, read the centre and spread. The distribution is centred on its mean np=10×0.4=4np = 10 \times 0.4 = 4, marked by the dashed line, and its spread is governed by the variance np(1−p)=10×0.4×0.6=2.4np(1 - p) = 10 \times 0.4 \times 0.6 = 2.4 (standard deviation 2.4≈1.55\sqrt{2.4} \approx 1.55). With p<12p < \tfrac{1}{2} the tail leans to the right, exactly as the skew rule predicts.

Binomial distribution B(10, 0.4)A bar chart of P(X equals k) for a binomial distribution with ten trials and success probability 0.4. The tallest bars are at k equals 4, near the mean of 4.kP(X = k)012345678910mean np = 4Stage 3The distribution is centred on the mean np = 4; spread is set by npq = 2.4.

How exam questions ask about the binomial distribution

The binomial is one of the most reliable sources of marks in Extension 1 statistics, and the question wordings are stable:

  • "Let XX be the number of [successes] in nn [trials] ... find E(X)E(X), Var(X)\text{Var}(X) and the standard deviation": identify nn and pp, then quote npnp, np(1−p)np(1 - p) and its square root. State the model X∼B(n,p)X \sim B(n, p) first.
  • "Find the probability of exactly kk [successes]": substitute into (nk)pk(1−p)n−k\binom{n}{k} p^k (1 - p)^{n - k} and evaluate. Showing the substituted formula earns the method mark even before you reach the calculator.
  • "Explain why XX is (or is not) binomial": name the four conditions, fixed nn, two outcomes, independence, constant pp, and say which one fails. Sampling without replacement is the classic failure (the model becomes hypergeometric).
  • "Let Y=aX+bY = aX + b ... find E(Y)E(Y) and Var(Y)\text{Var}(Y)": apply E(aX+b)=aE(X)+bE(aX + b) = aE(X) + b and Var(aX+b)=a2Var(X)\text{Var}(aX + b) = a^2 \text{Var}(X); the constant bb never touches the variance.
  • Distribution-shape parts: a sketch or a "describe the shape" cue wants you to mark the mean npnp and call the shape symmetric (if p=12p = \tfrac{1}{2}), right-skewed (p<12p < \tfrac{1}{2}) or left-skewed (p>12p > \tfrac{1}{2}).

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC Q213 marksA biased coin lands heads with probability 0.40.4. It is flipped 1010 times. Let XX be the number of heads. Find E(X)E(X), Var(X)\text{Var}(X) and the standard deviation.
Show worked answer →

X∼B(10,0.4)X \sim B(10, 0.4).

E(X)=np=10â‹…0.4=4E(X) = n p = 10 \cdot 0.4 = 4.

Var(X)=np(1−p)=10⋅0.4⋅0.6=2.4\text{Var}(X) = n p (1 - p) = 10 \cdot 0.4 \cdot 0.6 = 2.4.

Standard deviation: σ=2.4≈1.549\sigma = \sqrt{2.4} \approx 1.549.

Markers reward identifying the binomial model with n=10n = 10, p=0.4p = 0.4, applying the formulas, and the square root for σ\sigma.

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