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hscscience Ext 1 · Y12
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Module 10 · L18 of 20 ~40 min ⚡ +90 XP available

Assumptions and Limitations

A bag of $20$ marbles contains $8$ red. You draw $5$ marbles without replacement. Can you use the binomial distribution? A surveyor asks people in a queue whether they recycle, but the people are friends in a group. Can you use binomial here? These questions probe the four conditions the binomial requires — and what to do when one of them fails.

Today's hook — You shuffle a standard $52$-card deck and draw $5$ cards without replacement, counting hearts. Is this a binomial? If not, which condition fails, and what changes if you draw with replacement? Note your reasoning, then check after card 05.
0/5QUESTS
01
Recall — your gut answer first
+5 XP warm-up

From memory: list every condition you can think of that must hold for $X \sim \text{Bin}(n, p)$ to be the right model. Aim for at least three. Don't worry about precise wording yet.

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The four BINS conditions
+5 XP to read

The binomial distribution is only valid when all four conditions hold. A useful mnemonic is BINS: Binary outcomes, Independent trials, fixed Number of trials, and Same probability of success on each trial. If even one of these fails, the binomial formula gives wrong probabilities.

The BINS checklist for a binomial scenario:

B — Binary: each trial has exactly two outcomes (success / failure).
I — Independent: the outcome of one trial does not affect the others.
N — Fixed $n$: the number of trials is decided in advance.
S — Same $p$: the probability of success is identical on every trial.

B I N S All four must hold for Bin(n, p) to apply Any failure → binomial gives wrong answer
BINS → $X \sim \text{Bin}(n, p)$
Sampling without replacement
If a sample is drawn without replacement from a small population, trials are not independent and $p$ changes after each draw — both I and S fail. The correct model is the hypergeometric (out of Ext 1 scope).
Large-population rule
When the population is large (e.g. $> 10n$), sampling without replacement is approximately the same as with replacement, and the binomial is still a reasonable model. This is what HSC questions usually rely on.
"Number of trials until success"
If $n$ is not fixed — e.g. "the number of attempts until the first six" — N fails. That is a geometric distribution, not a binomial.
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What you'll master
Know

Key facts

  • The four BINS conditions: Binary, Independent, fixed $N$, same $p$
  • Why sampling without replacement breaks independence and changes $p$
  • The large-population rule: if pop $> 10n$, binomial is a good approximation
Understand

Concepts

  • Why all four conditions matter — failing any one invalidates the model
  • How to identify which condition fails in a real-world scenario
  • What alternative distributions apply (geometric, hypergeometric) when binomial fails
Can do

Skills

  • Test a scenario against all four BINS conditions, ticking off each one
  • Identify exactly which condition fails when a binomial isn't appropriate
  • Decide when the large-population approximation makes binomial usable anyway
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Key terms
Bernoulli trialA single trial with exactly two outcomes (success / failure) and constant success probability $p$. The binomial counts the successes in $n$ independent Bernoulli trials.
IndependenceTwo trials are independent if the outcome of one does not affect the probability of the other. Sampling with replacement preserves independence; without replacement breaks it.
Constant probabilityThe success probability $p$ must be the same on every trial. If the trials describe different scenarios (e.g. weather across different days), $p$ may vary and binomial fails.
Geometric distributionCounts the number of trials until the first success. $n$ is not fixed, so this is not binomial. Beyond Ext 1 scope but worth recognising.
Hypergeometric distributionModels successes when sampling without replacement from a finite population. The correct alternative to binomial in those cases; beyond Ext 1 scope.
ME12-5NESA outcome: applies appropriate statistical processes to present, analyse and interpret data, including identifying when a binomial model is appropriate.
05
Testing a scenario against BINS
core concept

For every binomial question, run a quick BINS check before writing any formula:

  1. B — Is each trial binary? Define "success" and "failure" clearly.
  2. I — Are the trials independent? Sampling with replacement, or from a large population, usually qualifies.
  3. N — Is $n$ fixed in advance? "Until the first success" means $n$ is random — geometric, not binomial.
  4. S — Is $p$ constant on every trial? Watch for scenarios where conditions change between trials.

Worked through the hook: Drawing $5$ cards without replacement, count hearts.

  • B ✓ — Each card is either a heart or not.
  • I ✗ — The first card's outcome changes the deck composition for the second draw.
  • N ✓ — $n = 5$ is fixed.
  • S ✗ — Initial $p = 13/52 = 1/4$, but after drawing a heart $p = 12/51 \neq 1/4$.
  • So without replacement, this is not binomial. With replacement, all four pass: $X \sim \text{Bin}(5, 0.25)$.
Practical rescue. If the population is much larger than the sample (rule of thumb: pop $> 10n$), the change in $p$ between draws is tiny and the binomial is still a very good approximation, even without replacement. That's why opinion polls of $n=1000$ from a population of millions are routinely modelled as binomial.

For every binomial question, run a quick BINS check before writing any formula:

Pause — copy the BINS checklist as four yes/no questions and work through the hook example to identify which letters pass into your book.

Quick check: A jar contains $10$ red and $10$ blue marbles. You draw $4$ marbles without replacement. Let $X$ be the number of red marbles drawn. Which BINS condition is the main reason this is not a true binomial?

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What goes wrong when conditions fail
core concept

We just saw the BINS checklist: B (binary outcomes), I (independent trials), N (fixed number of trials), S (same $p$). That raises a question: when one condition fails, what specific error does it introduce — and which alternative model becomes appropriate instead? This card answers it → I fails → hypergeometric model; N fails → geometric model; B fails → multinomial; S fails → no standard model.

Each failed condition corrupts the binomial in a specific, predictable way:

  • B fails (more than two outcomes): you cannot define "success" cleanly. Solution: regroup outcomes into two categories first (e.g. "rolled a six" vs "didn't roll a six").
  • I fails (trials affect each other): binomial overstates the variance. Use hypergeometric for finite-population sampling.
  • N fails ($n$ random): binomial cannot describe a random number of trials. If you're counting trials until the first success, use the geometric distribution.
  • S fails ($p$ varies): binomial assumes a single $p$. If $p$ differs across trials, you have a "Poisson binomial" — there is no clean formula and you usually must enumerate cases.
$$P(X = k) = \binom{n}{k} p^k q^{n-k} \quad \text{requires BINS all four to hold}$$
HSC strategy. Examiners often include a "discuss the assumptions" mark. State the four conditions explicitly, tick the ones that hold, and identify exactly which one(s) might be questionable. Vague answers like "it's binomial because there are two outcomes" lose marks — the marker wants all four checks.

Each failed condition corrupts the binomial in a specific, predictable way:

Pause — copy the four failure modes with their consequences: I fails (dependence), N fails (variable length), B fails (multi-outcome), S fails (changing $p$) into your book.

Did you get this? True or false: a die is rolled until a six appears for the first time, and $X$ is the number of rolls required. Then $X$ follows a binomial distribution.

PROBLEM 1 · CHECK ALL FOUR CONDITIONS

A free-throw shooter has a long-run success rate of $0.8$. She attempts $25$ free throws, and $X$ counts the successes. Justify whether $X \sim \text{Bin}(25, 0.8)$ is an appropriate model by checking each BINS condition.

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B: each shot is either made (success) or missed (failure) ✓.
Two outcomes per trial — the simplest condition to verify.
PROBLEM 2 · IDENTIFY THE FAILED CONDITION

A bag contains $6$ red and $4$ green balls. Three balls are drawn without replacement. Let $X$ be the number of red balls drawn. Explain why $X$ is not binomially distributed, and state which BINS condition(s) fail.

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B: red vs not-red — two outcomes ✓. N: $n = 3$ fixed ✓.
Two conditions pass straight away — define them carefully, then move to the harder ones.
PROBLEM 3 · LARGE-POPULATION RESCUE

A city of $2{,}000{,}000$ adults has support rate $0.4$ for a policy. A pollster samples $n = 800$ adults without replacement and counts supporters. Is the binomial model appropriate? Justify.

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B: support / don't support ✓. N: $n = 800$ fixed ✓.
Two clear ticks. The harder question is whether I and S survive the without-replacement sampling.

Fill the gap: The four BINS conditions for a binomial model are: outcomes, trials, fixed , and same probability $p$ on every trial.

Trap 01
Only mentioning "two outcomes"
"It's a binomial because each trial has two outcomes" is the most common under-answer. Markers want all four BINS conditions stated. Listing only one earns at most $1$ of usually $3$–$4$ marks for a "justify the model" question.
Trap 02
Calling sampling-without-replacement binomial
Students often default to binomial when sampling without replacement from a small population. Always check the pop/$n$ ratio. From a bag of $20$ marbles, $n=5$: ratio is $4 \ll 10$ — binomial is inappropriate. Identify I and S as the failed conditions.
Trap 03
Missing the "$n$ not fixed" case
"How many rolls until the first six?" or "how many shots until the first miss?" have a random $n$. The N condition fails. These are geometric distributions, not binomial. Always check: is the number of trials decided in advance, or determined by an outcome?

Did you get this? True or false: in a population of $50{,}000$ people, sampling $n = 100$ without replacement still gives an excellent binomial approximation because the population is far larger than the sample.

Work mode · how are you completing this lesson?
1

For each scenario, decide if a binomial model is appropriate and justify by listing each BINS condition: (i) flipping a fair coin $20$ times and counting heads; (ii) drawing $4$ cards from a $52$-card deck without replacement and counting aces.

2

Explain in one or two sentences why "the number of dice rolls until the first six" is not a binomial random variable. Which condition fails?

3

A pollster samples $200$ voters from a city of $1$ million without replacement. The true support rate for a candidate is $0.45$. Is binomial appropriate? Justify using the large-population rule.

4

A factory worker inspects $50$ items chosen from a batch of $80$, without replacement. The defective rate in the batch is $5\%$. Identify which BINS condition(s) fail and whether the large-population rule rescues the binomial.

5

An archer takes $10$ shots at a target. The probability of hitting the target increases by $0.02$ on each successive shot as she warms up (so $p_1 = 0.5, p_2 = 0.52, \dots, p_{10} = 0.68$). Is this a binomial situation? If not, which condition fails?

Odd one out: Three of these scenarios are valid binomial random variables. Which one is NOT?

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Revisit your thinking

Earlier you listed the conditions you thought the binomial required.

The complete list is BINS: Binary outcomes, Independent trials, fixed Number of trials, and Same probability of success on each trial. The most commonly missed conditions are I (independence) and S (constant $p$), which both fail in without-replacement sampling from small populations. Whenever you see "without replacement", run the pop/$n$ check.

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01
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.

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Short answer
ApplyBand 32 marks

Q1. State the four conditions (BINS) that must hold for a random variable $X$ to be modelled by $\text{Bin}(n, p)$. (2 marks)

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ApplyBand 43 marks

Q2. A bag contains $5$ red and $5$ blue marbles. Four marbles are drawn without replacement. Let $X$ be the number of red marbles drawn. Explain whether $X \sim \text{Bin}(4, 0.5)$, naming any condition(s) that fail. (3 marks)

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AnalyseBand 53 marks

Q3. A research team is studying flu transmission in classrooms of $25$ students. They model the number of infected students per classroom as $\text{Bin}(25, 0.1)$. Identify two reasons (i.e. two BINS conditions) that are likely violated in this setting, and explain how each one might fail. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. (i) All four BINS hold ($n=20$, $p=0.5$, independent flips, two outcomes) — binomial valid. (ii) Without replacement from a small deck: I and S fail (drawing one ace changes the deck and the probability for the next draw). Pop/n = 52/4 = 13 — borderline; binomial is approximate but not exact.

2. N fails. The number of rolls is random (depends on when the first six occurs), not fixed in advance. This is a geometric distribution, $\text{Geom}(1/6)$.

3. Pop/n = $10^6/200 = 5000 \gg 10$. Removing $200$ people from a million barely shifts $p$. I and S hold approximately, so $X \approx \text{Bin}(200, 0.45)$ is an excellent model.

4. Pop/n = $80/50 = 1.6 \ll 10$. Without replacement, I and S fail markedly. Binomial inappropriate — use hypergeometric for an exact answer.

5. S fails. $p$ varies across the $10$ shots ($p_1=0.5$ up to $p_{10}=0.68$). Not binomial. (This is a Poisson binomial, beyond Ext 1 scope, but the failure of S is the key observation.)

Q1 (2 marks): B – each trial has exactly two outcomes (success/failure); I – trials are independent; N – $n$ is fixed in advance; S – the success probability $p$ is the same on every trial [1 mark for at least 3, 2 marks for all 4 clearly stated].

Q2 (3 marks): Without replacement: I fails because each draw changes the bag composition [1]. S fails because $p$ changes from $0.5$ to either $4/9$ or $5/9$ after the first draw, etc. [1]. Pop/$n = 10/4 = 2.5 \ll 10$ so the large-population rescue does not apply — $X \not\sim \text{Bin}(4, 0.5)$ [1].

Q3 (3 marks): I fails: students in the same classroom interact, so once one student is infected the probability others become infected increases — trials are not independent [1.5]. S fails: students have different immunity levels (vaccination, prior infection, age), so $p$ is not constant across the class [1.5]. (Either I or S would also be acceptable; full marks for both clearly stated with reasoning.)

01
Boss battle · The BINS Inspector
earn bronze · silver · gold

Five timed scenario questions where you have to identify whether the binomial applies and, if not, which condition fails. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering BINS-check questions. Lighter alternative to the boss.

Mark lesson as complete

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