What is the binomial distribution, and what are its mean and variance?
Define the binomial distribution B(n,p), state its probability mass function, and find its mean np and variance np(1−p)
A focused answer to the HSC Maths Extension 1 dot point on the binomial distribution. The pmf P(X=k)=(kn​)pk(1−p)n−k, mean np, variance np(1−p), and standard situations that fit the model.
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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 X be the number of successes in n independent Bernoulli trials, each with success probability p. Then X has the binomial distribution B(n,p) (or Bin(n,p)).
The parameters are:
n: the number of trials (a positive integer).
p: the success probability on each trial (0≤p≤1).
q=1−p: the failure probability.
The probability mass function
For k=0,1,2,…,n,
P(X=k)=(kn​)pk(1−p)n−k.
The binomial coefficient (kn​) counts the number of ways to arrange k successes among n trials.
Mean and variance
By summing n independent Bernoulli trials (each with mean p and variance p(1−p)):
E(X)=np,Var(X)=np(1−p),σ=np(1−p)​.
Conditions for the binomial model
Four conditions must hold:
A fixed number of trials, n.
Each trial has exactly two outcomes (success or failure).
The trials are independent.
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 n coin flips: X∼B(n,p) with p the heads probability.
Number of items defective in a batch of n (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=21​, the distribution is symmetric around 2n​.
If p<21​, the distribution is right-skewed (long tail to the right).
If p>21​, it is left-skewed.
The skew decreases as n grows; for large n, 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=0.4 flipped n=10 times, so X∼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)=(k10​)(0.4)k(0.6)10−k. For the first few values, P(X=0)=(0.6)10≈0.006, P(X=1)=(110​)(0.4)(0.6)9≈0.040, and P(X=2)=(210​)(0.4)2(0.6)8≈0.121. Each is one number from the same formula.
Stage 2, fill in every k to get the whole distribution. Repeating the formula for k=0,1,…,10 gives all eleven bars. Because the eleven probabilities are the terms of (0.6+0.4)10, they sum to 110=1: the distribution is complete and valid.
Stage 3, read the centre and spread. The distribution is centred on its mean np=10×0.4=4, marked by the dashed line, and its spread is governed by the variance np(1−p)=10×0.4×0.6=2.4 (standard deviation 2.4​≈1.55). With p<21​ the tail leans to the right, exactly as the skew rule predicts.
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 X be the number of [successes] in n [trials] ... find E(X), Var(X) and the standard deviation": identify n and p, then quote np, np(1−p) and its square root. State the model X∼B(n,p) first.
"Find the probability of exactly k [successes]": substitute into (kn​)pk(1−p)n−k and evaluate. Showing the substituted formula earns the method mark even before you reach the calculator.
"Explain why X is (or is not) binomial": name the four conditions, fixed n, two outcomes, independence, constant p, and say which one fails. Sampling without replacement is the classic failure (the model becomes hypergeometric).
"Let Y=aX+b ... find E(Y) and Var(Y)": apply E(aX+b)=aE(X)+b and Var(aX+b)=a2Var(X); the constant b never touches the variance.
Distribution-shape parts: a sketch or a "describe the shape" cue wants you to mark the mean np and call the shape symmetric (if p=21​), right-skewed (p<21​) or left-skewed (p>21​).
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 0.4. It is flipped 10 times. Let X be the number of heads. Find E(X), Var(X) and the standard deviation.
Show worked answer →
X∼B(10,0.4).
E(X)=np=10â‹…0.4=4.
Var(X)=np(1−p)=10⋅0.4⋅0.6=2.4.
Standard deviation: σ=2.4​≈1.549.
Markers reward identifying the binomial model with n=10, p=0.4, applying the formulas, and the square root for σ.