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QLDMath MethodsSyllabus dot point

Topic 3: Discrete random variables

Recognise the binomial distribution XBin(n,p)X \sim \mathrm{Bin}(n, p) as the count of successes in nn independent Bernoulli trials, apply the binomial probability formula and use CAS, and use the formulas E(X)=npE(X) = np and Var(X)=np(1p)\mathrm{Var}(X) = np(1 - p)

A focused answer to the QCE Mathematical Methods Unit 3 dot point on the binomial distribution. Defines the binomial conditions (BINS), states the probability formula, gives the mean npnp and variance np(1p)np(1 - p), and walks through both by-hand Paper 1 calculations and CAS-supported Paper 2 calculations including P(Xk)P(X \leq k), P(Xk)P(X \geq k) and modelling applications.

Generated by Claude Opus 4.89 min answer

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What this dot point is asking

QCAA wants you to recognise when a count is binomially distributed, calculate binomial probabilities (by hand for small nn, by CAS for larger nn), and apply the mean and variance formulas. The binomial distribution is the most heavily examined probability model in QCE Mathematical Methods Unit 3, and it appears in IA1 PSMTs, IA2 short and extended response, and most EA Paper 2 probability questions.

The answer

When is XX binomial: the BINS conditions

A random variable XX has a binomial distribution if all four conditions hold.

  • B (Binary): Each trial has two outcomes labelled success or failure.
  • I (Independent): Trials are independent of one another.
  • N (Number fixed): The number of trials nn is fixed in advance.
  • S (Same probability): The probability of success pp is the same on every trial.

If all four hold, write XBin(n,p)X \sim \mathrm{Bin}(n, p), where XX is the number of successes in the nn trials.

The binomial probability formula

For XBin(n,p)X \sim \mathrm{Bin}(n, p) and k=0,1,2,,nk = 0, 1, 2, \ldots, n,

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

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

Paper 1 expects this formula by hand for small nn (typically n6n \leq 6) with values that simplify to clean fractions.

Paper 2 expects you to use CAS for any nn above about 6, calling functions named binomPdf(n,p,k)\mathrm{binomPdf}(n, p, k) for P(X=k)P(X = k) and binomCdf(n,p,a,b)\mathrm{binomCdf}(n, p, a, b) for P(aXb)P(a \leq X \leq b). Check the exact syntax for your approved CAS model.

Mean and variance

For XBin(n,p)X \sim \mathrm{Bin}(n, p),

E(X)=np,Var(X)=np(1p).E(X) = n p, \qquad \mathrm{Var}(X) = n p (1 - p).

Standard deviation: σ=np(1p)\sigma = \sqrt{n p (1 - p)}.

These come from the fact that a binomial is the sum of nn independent Bernoulli trials, each with mean pp and variance p(1p)p (1 - p), and the rules E(X+Y)=E(X)+E(Y)E(X + Y) = E(X) + E(Y) and Var(X+Y)=Var(X)+Var(Y)\mathrm{Var}(X + Y) = \mathrm{Var}(X) + \mathrm{Var}(Y) for independent variables.

Cumulative probabilities

QCAA commonly asks for P(Xk)P(X \leq k), P(Xk)P(X \geq k) or P(aXb)P(a \leq X \leq b) rather than a single P(X=k)P(X = k).

  • P(Xk)=i=0kP(X=i)P(X \leq k) = \sum_{i = 0}^{k} P(X = i).
  • P(Xk)=1P(Xk1)P(X \geq k) = 1 - P(X \leq k - 1) (use the complement to save work).
  • P(aXb)=P(Xb)P(Xa1)P(a \leq X \leq b) = P(X \leq b) - P(X \leq a - 1).

For nn above about 6, do all four with CAS. For Paper 1 with small nn, set up the sums explicitly and evaluate by hand.

Exam-style practice questions

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

2023 QCAA-style P13 marksA fair six-sided die is rolled 5 times. Let XX be the number of sixes. Find P(X=2)P(X = 2) exactly.
Show worked answer →

XBin(5,16)X \sim \mathrm{Bin}(5, \frac{1}{6}) because rolling a six is a single success in each independent trial.

P(X=2)=(52)(16)2(56)3=10136125216=12507776=6253888.P(X = 2) = \binom{5}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^3 = 10 \cdot \frac{1}{36} \cdot \frac{125}{216} = \frac{1250}{7776} = \frac{625}{3888}.

As a decimal, 0.1608\approx 0.1608.

Markers reward the correct identification of n=5n = 5 and p=16p = \frac{1}{6}, the binomial coefficient (52)=10\binom{5}{2} = 10, the powers of pp and 1p1 - p matched to successes and failures, and the simplified exact fraction. Paper 1 wants exact form.

2022 QCAA-style P25 marksA quality control process tests batches of 20 components from a production line. The probability that any one component is faulty is 0.050.05, independently of the others. Let XX be the number of faulty components in a batch. (a) Find P(X=0)P(X = 0). (b) Find P(X2)P(X \geq 2). (c) Find the expected number of faulty components per batch and the standard deviation.
Show worked answer →

A 5-mark answer needs the distribution statement, both probabilities (with the complement trick for part (b)), and the descriptive statistics.

XBin(20,0.05)X \sim \mathrm{Bin}(20, 0.05).

(a) P(X=0)=(0.95)200.3585P(X = 0) = (0.95)^{20} \approx 0.3585.

(b) Use the complement: P(X2)=1P(X=0)P(X=1)P(X \geq 2) = 1 - P(X = 0) - P(X = 1).

P(X=1)=(201)(0.05)(0.95)19=200.05(0.95)190.3774P(X = 1) = \binom{20}{1} (0.05)(0.95)^{19} = 20 \cdot 0.05 \cdot (0.95)^{19} \approx 0.3774.

P(X2)=10.35850.37740.2641P(X \geq 2) = 1 - 0.3585 - 0.3774 \approx 0.2641.

(CAS: 1binomCdf(20,0.05,0,1)0.26411 - \mathrm{binomCdf}(20, 0.05, 0, 1) \approx 0.2641.)

(c) E(X)=np=200.05=1E(X) = n p = 20 \cdot 0.05 = 1. Var(X)=np(1p)=200.050.95=0.95\mathrm{Var}(X) = n p (1 - p) = 20 \cdot 0.05 \cdot 0.95 = 0.95. Standard deviation 0.9747\approx 0.9747.

Markers reward stating the distribution explicitly, the complement-trick setup for P(X2)P(X \geq 2) (rather than summing 19 terms), the CAS-friendly numerical answers to 4 decimal places, and the mean-variance formulas applied cleanly.

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