Your weak spots

Insights load after your first practice round.

L20 of 20 · Final ~45 min ⚡ +100 XP available

Module 10 Synthesis & Exam Technique

You have arrived at the final lesson of HSC Mathematics Extension 1. The HSC exam doesn't tell you which technique to use — it tests whether you can write $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$ explicitly, show all working for method marks, recognise when to apply the complement, and know when a normal approximation is warranted. This lesson consolidates every formula from Module 10 and equips you with the exam-room habits that separate Band 5 from Band 6.

Today's hook — Before reading on, write down — from memory — the three core binomial formulas: $P(X = k)$, $E(X)$, and $\operatorname{Var}(X)$. Then state the condition under which a normal approximation to the binomial is appropriate. Compare your answers after card 05.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

Across Module 10 you learnt the binomial PMF, the formulas for mean and variance, and the normal approximation. Before turning the page — write down every formula you can recall, including the condition for a normal approximation. This is your end-of-course memory audit.

auto-saved

The two moves for exam-room success

+5 XP to read

Every HSC binomial question rewards two habits: write $P(X = k)$ explicitly before substituting, and show every line of working so the marker awards method marks even if your arithmetic slips. Skipping working to "save time" routinely loses 1–2 marks per question.

The explicit-and-show strategy: (1) state the distribution $X \sim B(n, p)$, (2) write the formula, e.g. $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$, before any substitution, (3) substitute numerically and evaluate, (4) write a final-answer line with units / context.

$E(X) = np$ · $\operatorname{Var}(X) = np(1-p)$ · $\sigma = \sqrt{np(1-p)}$

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

Write the formula first

Every binomial answer should begin with $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$ before substitution. Markers explicitly look for this line.

Use the complement

For "at least one" use $1 - (1-p)^n$. For "at most $k$" with large $k$ use $1 - P(X \geq k+1)$. The complement is a marker of fluency.

Approximate when valid

If $np \geq 5$ and $n(1-p) \geq 5$, the binomial can be approximated by $N(np, np(1-p))$ — useful for cumulative probabilities with large $n$.

What you'll master

Know

Key facts

All Module 10 formulas: $P(X = k)$, $E(X)$, $\operatorname{Var}(X)$, $\sigma$
Complement identities: $P(X \geq 1) = 1 - (1-p)^n$
Normal approximation condition: $np \geq 5$ and $n(1-p) \geq 5$

Understand

Concepts

Why writing $P(X = k)$ explicitly secures method marks
How the complement avoids long summations
When normal approximation is appropriate and when to use the exact binomial

Can do

Skills

Tackle any HSC Module 10 question with the full four-step layout
Choose between exact binomial and normal approximation appropriately
Show every line of working to maximise method marks

Key terms

Binomial distribution$X \sim B(n, p)$ counts the number of successes in $n$ independent Bernoulli trials. Discrete; supported on $\{0, 1, \ldots, n\}$.

PMFProbability mass function. For binomial: $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$. Always write this line in your solution.

Method marksMarks awarded for writing the correct setup or formula, even if the final numerical answer is wrong. Show every step to claim them.

Complement$P(A) = 1 - P(\text{not } A)$. The single most useful trick for "at least one" and large-$k$ "at most" questions.

Normal approximationWhen $np \geq 5$ and $n(1-p) \geq 5$, $B(n, p) \approx N(np, np(1-p))$. Standardise $Z = (X - np)/\sqrt{np(1-p)}$ to use the standard normal table.

ME12-5NESA outcome: applies appropriate statistical processes to present, analyse and interpret data, including the binomial distribution and its approximations.

The complete Module 10 formula sheet

core concept

Memorise these — they are the entire algebraic content of Module 10:

PMF: $P(X = k) = \displaystyle\binom{n}{k}p^k(1-p)^{n-k}$ for $k = 0, 1, \ldots, n$.
Mean: $E(X) = np$.
Variance: $\operatorname{Var}(X) = np(1-p)$.
Standard deviation: $\sigma = \sqrt{np(1-p)}$.
Complement (at least one): $P(X \geq 1) = 1 - (1-p)^n$.
Normal approximation: if $np \geq 5$ and $n(1-p) \geq 5$, then $X \approx N(np, np(1-p))$ and $Z = \dfrac{X - np}{\sqrt{np(1-p)}}$ is approximately $N(0, 1)$.

Worked through the hook: For $X \sim B(40, 0.4)$ (a typical HSC scenario):

$E(X) = 16$, $\operatorname{Var}(X) = 9.6$, $\sigma \approx 3.10$.
$np = 16 \geq 5$ and $n(1-p) = 24 \geq 5$, so normal approximation is valid.
$P(X \geq 20) \approx P\left(Z \geq \dfrac{20 - 16}{\sqrt{9.6}}\right) = P(Z \geq 1.29) \approx 0.0985$.

When to approximate vs use exact binomial. If the question gives small $n$ (say $n \leq 20$) and asks for $P(X = k)$, use the exact binomial. If $n$ is large (e.g. $n = 100$) and the question asks for a cumulative probability like $P(X \geq 60)$, the normal approximation is far more practical — and the markers expect it.

Memorise these — they are the entire algebraic content of Module 10:

Pause — copy the complete Module 10 formula sheet: PMF, $E(X)=np$, $\text{Var}(X)=npq$, $\text{SD}(X)=\sqrt{npq}$, complement, and normal approximation condition into your book.

Quick check: Under which condition is it appropriate to approximate $B(n, p)$ by a normal distribution?

Exam-room habits that bank marks

core concept

We just saw the complete Module 10 formula sheet: PMF, $E(X)=np$, $\text{Var}(X)=npq$, SD, complement shortcut, and normal approximation conditions. That raises a question: beyond knowing the formulas, what five exam-room habits ensure you earn every available mark even when the arithmetic is correct? This card answers it → state the distribution, write the full formula before substituting, show intermediate calculations, round correctly, and write a concluding sentence.

HSC markers use a marking rubric. Each step in your solution is graded against that rubric, so the layout of your answer matters as much as the arithmetic. Five habits to drill:

State the distribution. "Let $X$ be the number of successes; $X \sim B(n, p)$." This earns a method mark on its own.
Quote the formula. Write $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$ before substituting.
Show substitution. Write the numerical version, e.g. $\binom{20}{6}(0.3)^6(0.7)^{14}$, before evaluating.
Use the complement explicitly. Write "$P(X \geq 1) = 1 - P(X = 0)$" — don't just write the final number.
State the final answer with units. "The probability is approximately $0.1916$" — not just "$0.1916$".

$$P(X \geq k) = 1 - P(X \leq k-1) \quad \text{and} \quad P(X > k) = 1 - P(X \leq k)$$

Most common exam errors. (1) writing $p^{n-k}(1-p)^k$ instead of $p^k(1-p)^{n-k}$; (2) confusing variance with standard deviation; (3) trying to sum $P(X = 1) + \ldots + P(X = n)$ instead of using the complement; (4) using the normal approximation when $np < 5$.

HSC markers use a marking rubric. Each step in your solution is graded against that rubric, so the layout of your answer matters as much as the arithmetic. Five habits to drill:

Pause — copy the five exam habits: (1) state $X\sim B(n,p)$, (2) write formula before substituting, (3) show intermediate steps, (4) round to 4 d.p., (5) write conclusion into your book.

Did you get this? True or false: writing the binomial formula $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$ before substituting numbers can earn a method mark even if your final answer is incorrect.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · FULL FOUR-STEP LAYOUT

A coin biased so that $P(\text{head}) = 0.3$ is tossed 18 times. Let $X$ be the number of heads. (a) Find $E(X)$ and $\operatorname{Var}(X)$. (b) Find $P(X = 5)$.

Let $X$ be the number of heads; $X \sim B(18, 0.3)$. So $E(X) = np = 18 \times 0.3 = 5.4$ and $\operatorname{Var}(X) = np(1-p) = 18 \times 0.3 \times 0.7 = 3.78$.

State the distribution explicitly — this is a method mark on its own. Then quote the formula and substitute.

PROBLEM 2 · COMPLEMENT IN ACTION

A quality-control engineer inspects batches of 50 components. Each component independently has a 4% chance of being defective. Find the probability that a batch contains at least one defective component.

Let $X$ be the number of defectives; $X \sim B(50, 0.04)$.

State the distribution. $n = 50$, $p = 0.04$.

PROBLEM 3 · NORMAL APPROXIMATION

A multiple-choice test has 100 questions, each with 4 options. A student guesses every answer. Use a normal approximation to estimate $P(X \geq 30)$, the probability the student scores at least 30 correct.

$X \sim B(100, 0.25)$. Check the approximation condition: $np = 25 \geq 5$ and $n(1-p) = 75 \geq 5$ ✓. So $X \approx N(\mu, \sigma^2)$ with $\mu = np = 25$ and $\sigma^2 = np(1-p) = 18.75$, so $\sigma \approx 4.33$.

Always verify the $np, n(1-p) \geq 5$ condition before applying the approximation — markers explicitly check this step.

Fill the gap: For $X \sim B(100, 0.25)$, the normal approximation gives $\mu = 25$ and $\sigma^2 = np(1-p) = 100 \times 0.25 \times = 18.75$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Skipping the formula line

Students often write only the final number, e.g. "$P(X = 5) = 0.2017$". The marker has nothing to grade if the working is missing. Always write $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$ before substituting — it earns its own method mark and protects you against arithmetic slips.

Trap 02

Using normal approximation when invalid

If $np < 5$ or $n(1-p) < 5$, the binomial distribution is too skewed for a normal approximation to be reliable. Example: $X \sim B(20, 0.1)$ has $np = 2 < 5$, so use the exact binomial. The approximation gives wrong answers and the markers will deduct.

Trap 03

Confusing $\operatorname{Var}(X)$ with $\sigma$

$\operatorname{Var}(X) = np(1-p)$ — a squared quantity. $\sigma = \sqrt{np(1-p)}$ — the square root. When standardising, use $\sigma$ (not the variance) in the denominator: $Z = (X - np)/\sigma$. Plugging in the variance instead of the SD is one of the most common normal-approximation errors.

Did you get this? True or false: for $X \sim B(20, 0.1)$, it is appropriate to use a normal approximation because $n = 20$ is reasonably large.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A biased coin has $P(\text{head}) = 0.4$ and is tossed 25 times. Write down (a) $E(X)$ and (b) $\operatorname{Var}(X)$, and (c) the formula for $P(X = 10)$ (do not evaluate).

A 20% of voters favour a proposal. A random sample of 12 voters is taken. Find $P(X = 3)$ to 4 d.p., showing every line of working.

A drug is effective in 90% of patients. It is given to 8 patients. Find the probability of at least one failure, using the complement.

A factory produces 200 items per day, each with a 6% chance of being defective. Using a normal approximation, estimate the probability of more than 15 defectives. Justify the use of the approximation.

An exam has 50 MC questions, 4 options each. A student guesses every answer. Find (a) the expected score, (b) the standard deviation, and (c) $P(X \geq 20)$ using the normal approximation.

Odd one out: Three of these are valid Module 10 strategies for an HSC exam answer. Which one is NOT a recommended habit?

Revisit your thinking · course complete

At the start of this lesson you wrote down — from memory — the binomial formulas and the normal approximation condition.

The full sheet: $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$, $E(X) = np$, $\operatorname{Var}(X) = np(1-p)$, $\sigma = \sqrt{np(1-p)}$, $P(X \geq 1) = 1 - (1-p)^n$, and normal approximation valid when $np \geq 5$ and $n(1-p) \geq 5$. If any of these were missing from your initial recall, drill them now — these are the six lines that unlock every HSC binomial question.

This is the final lesson of HSC Mathematics Extension 1. Congratulations on completing the course. Treat every binomial question with the four-step layout: state the distribution, quote the formula, substitute, finalise. Method marks are yours for the taking.

auto-saved

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Let $X \sim B(30, 0.2)$. Find $E(X)$, $\operatorname{Var}(X)$ and $\sigma$. (2 marks)

auto-saved

ApplyBand 43 marks

Q2. A pharmaceutical company tests a new vaccine on 15 patients, each with a 0.85 probability of seroconversion. Find $P(X = 13)$ to 4 d.p. and $P(X \geq 1)$, showing every step of working. (3 marks)

auto-saved

AnalyseBand 53 marks

Q3. A factory produces 500 light globes per hour. Each independently has a 3% chance of being defective. Using a normal approximation, estimate the probability that an hour's output contains fewer than 10 defectives. Justify the use of the approximation. (3 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:

1. $X \sim B(25, 0.4)$: $E(X) = 10$, $\operatorname{Var}(X) = 25 \times 0.4 \times 0.6 = 6$, $P(X = 10) = \binom{25}{10}(0.4)^{10}(0.6)^{15}$.

2. $X \sim B(12, 0.2)$: $P(X = 3) = \binom{12}{3}(0.2)^3(0.8)^9 = 220 \times 0.008 \times 0.1342 \approx 0.2362$.

3. $X = $ number of failures, $X \sim B(8, 0.1)$: $P(X \geq 1) = 1 - (0.9)^8 = 1 - 0.4305 \approx 0.5695$.

4. $X \sim B(200, 0.06)$: $np = 12, n(1-p) = 188$, both $\geq 5$ ✓. $\mu = 12$, $\sigma^2 = 11.28$, $\sigma \approx 3.36$. $P(X > 15) \approx P(Z > (15-12)/3.36) = P(Z > 0.89) \approx 1 - 0.8133 = 0.1867$.

5. $X \sim B(50, 0.25)$: (a) $E(X) = 12.5$; (b) $\sigma = \sqrt{9.375} \approx 3.06$; (c) $P(X \geq 20) \approx P(Z \geq 2.45) \approx 1 - 0.9929 = 0.0071$.

Q1 (2 marks): $X \sim B(30, 0.2)$ [0.5]. $E(X) = 6$, $\operatorname{Var}(X) = 30 \times 0.2 \times 0.8 = 4.8$, $\sigma = \sqrt{4.8} \approx 2.19$ [1.5 — half-mark per value].

Q2 (3 marks): $X \sim B(15, 0.85)$ [1]. $P(X=13) = \binom{15}{13}(0.85)^{13}(0.15)^2 = 105 \times 0.1209 \times 0.0225 \approx 0.2856$ [1]. $P(X \geq 1) = 1 - (0.15)^{15} \approx 1 - 4.4\times10^{-13} \approx 1.0000$ [1].

Q3 (3 marks): $X \sim B(500, 0.03)$, $np = 15 \geq 5, n(1-p) = 485 \geq 5$ ✓ [1]. $\mu = 15, \sigma^2 = 14.55, \sigma \approx 3.81$ [1]. $P(X < 10) \approx P(Z < (10-15)/3.81) = P(Z < -1.31) = 1 - \Phi(1.31) \approx 1 - 0.9049 = 0.0951$ [1].

Boss battle · The Final Boss

earn bronze · silver · gold

Five timed HSC-style questions spanning every Module 10 technique — binomial PMF, mean and variance, complement, and normal approximation. The final boss of the entire HSC Extension 1 course. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering exam-style binomial questions. Lighter alternative to the boss.

Mark lesson as complete · course finished

Tick when you've finished the practice and review. This is the final lesson of HSC Mathematics Extension 1.

← Lesson 19 · Mixed Statistical Problems Module 10 overview →

Module overview · Extension 1 · Checkpoint 4