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Module 10 · L19 of 20 ~40 min ⚡ +90 XP available

Mixed Statistical Problems

A pharmaceutical company tests a new drug on 20 patients, knowing each independently has a 30% chance of remission. The clinical team must answer: what is the expected number of successes, the variance of the outcome, and the probability that at least 8 patients improve? Real HSC questions chain these calculations together. This lesson trains the multi-technique habit — moving fluently between $E(X)$, $\operatorname{Var}(X)$ and $P(X = k)$ in a single solution.

Today's hook — Let $X \sim B(20, 0.3)$. Before reading on, write down (a) $E(X)$, (b) $\operatorname{Var}(X)$, and (c) the formula you would use to compute $P(X = 6)$. Compare your answers after card 05.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

Consider a binomial random variable $X \sim B(n, p)$. Before using any formula — write the three core results: $P(X=k) = \ldots$, $E(X) = \ldots$, $\operatorname{Var}(X) = \ldots$. Sketch your reasoning below.

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The two moves for mixed binomial problems

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Every mixed statistics problem rewards two habits: identify $n$ and $p$ precisely (the Bernoulli parameters), then match the question to the correct formula — $E(X)$, $\operatorname{Var}(X)$, or $P(X=k)$ — before evaluating. Rushing to a calculator without checking which result is asked is the single biggest error in multi-part questions.

The identify-and-match strategy: (1) read the wording to find $n$ (number of trials) and $p$ (probability of success), (2) classify the sub-question as mean, variance, or probability, (3) write the formula in full before substituting.

$E(X) = np$ · $\operatorname{Var}(X) = np(1-p)$ · $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$

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

Check independence + fixed $n$

Binomial only applies when trials are independent, the number $n$ is fixed in advance, and $p$ is constant. If any of these break, the formulas above are invalid.

Complement trick

For "at least one" or "at most $k$", use $P(X \geq 1) = 1 - P(X = 0)$ or $P(X > k) = 1 - P(X \leq k)$. Avoids summing many terms.

$\sigma = \sqrt{np(1-p)}$

Standard deviation is the square root of variance. Don't confuse $\operatorname{Var}(X) = np(1-p)$ with $\sigma = \sqrt{np(1-p)}$ — examiners often ask for one and reward the other only with method marks.

What you'll master

Know

Key facts

Binomial PMF: $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$
Mean: $E(X) = np$ and Variance: $\operatorname{Var}(X) = np(1-p)$
Complement: $P(X \geq 1) = 1 - P(X = 0) = 1 - (1-p)^n$

Understand

Concepts

Why $n$ and $p$ must be identified before any computation
How mean, variance and probability connect through the same parameters
Why the complement is preferred for "at least one" questions

Can do

Skills

Solve multi-part HSC questions combining $E(X)$, $\operatorname{Var}(X)$ and $P(X = k)$
Use the complement to simplify "at least" and "at most" expressions
Apply the general strategy: read, classify, compute, verify

Key terms

Bernoulli trialA single trial with exactly two outcomes — success (probability $p$) or failure (probability $1-p$). Binomial counts the successes in $n$ independent Bernoulli trials.

Binomial PMF$P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$ — the probability of exactly $k$ successes in $n$ trials.

Expected valueFor $X \sim B(n, p)$: $E(X) = np$. The long-run average number of successes per $n$-trial experiment.

VarianceFor $X \sim B(n, p)$: $\operatorname{Var}(X) = np(1-p)$. Maximised when $p = 0.5$. Standard deviation $\sigma = \sqrt{np(1-p)}$.

Complement event$P(A) = 1 - P(\text{not } A)$. Used to convert "at least one" into $1 - P(\text{none})$, a single-term calculation.

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

The multi-part problem strategy

core concept

HSC binomial questions almost always combine two or three sub-parts in one question. The four-step strategy never fails:

Read. Extract $n$ (number of trials) and $p$ (probability of success) from the wording.
Classify. For each sub-part, identify: mean ($E$), variance ($\operatorname{Var}$), or probability ($P$).
Compute. Write the formula explicitly, then substitute. Never skip writing $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$.
Verify. Check that probabilities lie in $[0, 1]$, $E(X)$ is between $0$ and $n$, and $\operatorname{Var}(X) \geq 0$.

Worked through the hook: For $X \sim B(20, 0.3)$:

$E(X) = np = 20 \times 0.3 = 6$.
$\operatorname{Var}(X) = np(1-p) = 20 \times 0.3 \times 0.7 = 4.2$.
$P(X = 6) = \displaystyle\binom{20}{6}(0.3)^6(0.7)^{14} = 38760 \times (0.3)^6 \times (0.7)^{14} \approx 0.1916$.

Connecting the three results. A mixed problem might also ask you to compare the observed result (say, $X = 10$) to the mean: $10 - 6 = 4$, which is $4/\sqrt{4.2} \approx 1.95$ standard deviations above the mean. That puts it almost two SDs out — a useful "unlikely but not impossible" signal.

HSC binomial questions almost always combine two or three sub-parts in one question. The four-step strategy never fails:

Pause — copy the four-step multi-part strategy (read, classify, compute, verify) and identify which formula to apply for mean, variance, and probability sub-parts into your book.

Quick check: Let $X \sim B(15, 0.4)$. Which expression correctly gives $P(X = 5)$?

Combining mean, variance and probability in one problem

core concept

We just saw the four-step multi-part strategy: read (extract $n,p$), classify each sub-part as mean/variance/probability, write the formula, verify. That raises a question: when HSC questions chain mean, variance, and probability sub-parts together, how do you avoid re-reading parameters each time? This card answers it → extract $n$ and $p$ once at the top, reuse them in all three formulas ($E=np$, $\text{Var}=npq$, PMF) without re-defining variables.

HSC questions often chain three sub-parts: find the mean, find the variance, then compute a specific probability. Because all three depend on the same $n$ and $p$, you only need to read the parameters once and reuse them.

Example: A fair die is rolled 12 times. Let $X$ be the number of sixes. Here $X \sim B(12, 1/6)$.

Mean: $E(X) = np = 12 \times \tfrac{1}{6} = 2$.
Variance: $\operatorname{Var}(X) = np(1-p) = 12 \times \tfrac{1}{6} \times \tfrac{5}{6} = \tfrac{10}{6} = \tfrac{5}{3}$.
$P(X \geq 1)$: using the complement, $P(X \geq 1) = 1 - P(X = 0) = 1 - \left(\tfrac{5}{6}\right)^{12} \approx 0.8878$.

$$P(X \geq 1) = 1 - (1-p)^n$$

Common mistake. Students try to compute $P(X \geq 1)$ by summing $P(X = 1) + P(X = 2) + \ldots + P(X = n)$ — a 12-term sum here. The complement turns it into a single calculation: $1 - (5/6)^{12}$. Always use the complement when the question asks for "at least one".

Pause — copy the three chained formulas ($E(X)=np$, $\text{Var}(X)=npq$, $P(X=k)=\binom{n}{k}p^k q^{n-k}$) with a worked three-part example into your book.

Did you get this? True or false: for $X \sim B(12, 1/6)$, the probability of obtaining at least one six in 12 rolls is $1 - (5/6)^{12}$.

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

PROBLEM 1 · MEAN + VARIANCE + EXACT PROBABILITY

A multiple-choice test has 20 questions, each with 4 options and exactly one correct answer. A student guesses every answer. (a) Find $E(X)$ and $\operatorname{Var}(X)$ for the number of correct answers. (b) Find $P(X = 7)$ to 4 d.p.

$X \sim B(20, 0.25)$. So $E(X) = np = 20 \times 0.25 = 5$ and $\operatorname{Var}(X) = np(1-p) = 20 \times 0.25 \times 0.75 = 3.75$.

Always identify $n$ and $p$ first. Random guessing on 4-option MC gives $p = 1/4 = 0.25$.

PROBLEM 2 · COMPLEMENT + AT LEAST ONE

A factory produces light globes; 5% are defective. A batch of 25 is inspected. (a) Find the expected number of defectives. (b) Find the probability that the batch contains at least one defective globe.

$X \sim B(25, 0.05)$. $E(X) = np = 25 \times 0.05 = 1.25$.

Even with a small $p$, $E(X)$ is non-zero — expect around 1 to 2 defectives per batch.

PROBLEM 3 · CUMULATIVE PROBABILITY (P(X ≤ k))

A basketball player has a free-throw success rate of $p = 0.7$. She takes 10 free throws. Find $P(X \leq 2)$, the probability she sinks at most 2 baskets.

$X \sim B(10, 0.7)$. $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$.

For "at most $k$", sum from $0$ to $k$ when $k$ is small. (For large $k$, use the complement: $P(X \leq k) = 1 - P(X \geq k+1)$.)

Fill the gap: For $X \sim B(50, 0.4)$, the variance is $\operatorname{Var}(X) = 50 \times 0.4 \times = 12$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Swapping the exponents of $p$ and $1-p$

The formula is $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$: the exponent of $p$ matches the number of successes $k$, and the exponent of $1-p$ matches the failures $n-k$. Students often write $p^{n-k}(1-p)^k$ by mistake. Always double-check which exponent goes with $p$.

Trap 02

Using $\sigma$ when the question asks for $\operatorname{Var}(X)$ (or vice versa)

$\operatorname{Var}(X) = np(1-p)$ but $\sigma = \sqrt{np(1-p)}$. Read the question carefully — "find the variance" and "find the standard deviation" are different. Mixing the two loses easy marks even when the calculation is otherwise correct.

Trap 03

Forgetting the complement for "at least one"

"At least one" virtually always means use $P(X \geq 1) = 1 - P(X = 0)$. Trying to sum $P(X = 1) + P(X = 2) + \ldots + P(X = n)$ is impractical — for $n = 50$ that's 50 terms. The complement collapses it to one term: $1 - (1-p)^n$.

Did you get this? True or false: for $X \sim B(n, p)$, the standard deviation is given by $\sigma = np(1-p)$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A coin is biased so that $P(\text{head}) = 0.6$. It is tossed 10 times. Find $E(X)$ and $\operatorname{Var}(X)$ where $X$ is the number of heads.

In a town, 8% of people are left-handed. A random sample of 30 people is taken. Find the probability that exactly 2 are left-handed.

A drug cures 80% of patients. It is given to 15 patients. Find the probability that at least 14 are cured.

A factory has a 2% defect rate. A box of 100 items is inspected. Find the probability of at least one defective item.

In a class of 20 students, each independently has a 0.4 probability of completing homework. Find (a) the expected number, (b) the standard deviation, and (c) $P(X = 8)$.

Odd one out: Three of these statements about $X \sim B(20, 0.3)$ are correct. Which one is NOT?

Revisit your thinking

Earlier you wrote out $E(X)$, $\operatorname{Var}(X)$ and the $P(X = 6)$ setup for $X \sim B(20, 0.3)$.

The expected value is $E(X) = np = 6$. Variance is $\operatorname{Var}(X) = np(1-p) = 4.2$. The exact probability is $P(X = 6) = \binom{20}{6}(0.3)^{6}(0.7)^{14} \approx 0.1916$. Be especially careful that the exponent of $p$ matches the number of successes — swapping the exponents is the single most common error.

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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. A fair coin is tossed 8 times. Let $X$ be the number of heads. Find $E(X)$ and $\operatorname{Var}(X)$. (2 marks)

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ApplyBand 43 marks

Q2. A pharmaceutical trial finds that a drug works on 60% of patients. It is given to 12 patients. Find (a) $P(X = 7)$ and (b) $P(X \geq 1)$. Leave answers to 4 d.p. (3 marks)

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AnalyseBand 53 marks

Q3. A test has 25 multiple-choice questions, each with 5 options. A student guesses every answer. Find the expected score, the variance, and $P(X \geq 10)$ using the binomial formula. (You may leave $P(X \geq 10)$ as a sum of terms.) (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $X \sim B(10, 0.6)$: $E(X) = 6$, $\operatorname{Var}(X) = 10 \times 0.6 \times 0.4 = 2.4$.

2. $X \sim B(30, 0.08)$: $P(X=2) = \binom{30}{2}(0.08)^2(0.92)^{28} = 435 \times 0.0064 \times 0.0987 \approx 0.2745$.

3. $X \sim B(15, 0.8)$: $P(X \geq 14) = P(X=14) + P(X=15) = \binom{15}{14}(0.8)^{14}(0.2) + (0.8)^{15} = 15(0.8)^{14}(0.2) + (0.8)^{15} \approx 0.1319 + 0.0352 = 0.1671$.

4. $X \sim B(100, 0.02)$: $P(X \geq 1) = 1 - (0.98)^{100} \approx 1 - 0.1326 = 0.8674$.

5. $X \sim B(20, 0.4)$: (a) $E(X) = 8$; (b) $\operatorname{Var}(X) = 4.8$, $\sigma \approx 2.19$; (c) $P(X=8) = \binom{20}{8}(0.4)^8(0.6)^{12} = 125970 \times 6.554\times10^{-4} \times 2.177\times10^{-3} \approx 0.1797$.

Q1 (2 marks): $X \sim B(8, 0.5)$ [1]. $E(X) = 4$, $\operatorname{Var}(X) = 2$ [1].

Q2 (3 marks): $X \sim B(12, 0.6)$ [1]. (a) $P(X=7) = \binom{12}{7}(0.6)^7(0.4)^5 = 792 \times 0.02799 \times 0.01024 \approx 0.2270$ [1]. (b) $P(X \geq 1) = 1 - (0.4)^{12} \approx 1 - 0.0000168 \approx 1.0000$ [1].

Q3 (3 marks): $X \sim B(25, 0.2)$ [1]. $E(X) = 5$, $\operatorname{Var}(X) = 25 \times 0.2 \times 0.8 = 4$ [1]. $P(X \geq 10) = \sum_{k=10}^{25} \binom{25}{k}(0.2)^k(0.8)^{25-k} \approx 0.0173$ [1] (evaluating each term separately; some answers may also express it as $1 - P(X \leq 9)$).

Boss battle · The Binomial Combiner

earn bronze · silver · gold

Five timed questions combining $E(X)$, $\operatorname{Var}(X)$ and binomial probabilities. 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 mixed binomial questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

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