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Module 10 · L03 of 20 ~35 min ⚡ +90 XP available

Bernoulli Trials

A quality control engineer tests 12 light bulbs from a production line; each bulb is either defective or not. A student takes a 20-question multiple-choice quiz with four options per question, guessing randomly. Both situations share an identical structure — a fixed number of independent trials, each with the same two possible outcomes and the same probability of success. That structure is called a Bernoulli process, and it is the foundation of the binomial distribution.

Today's hook — You roll a fair six-sided die 10 times and record whether each roll is a six. Before reading on, write down: (1) what counts as "success" here, (2) the probability of success on any single roll, and (3) is the probability of success the same on every roll? Why or why not? Compare your answer to the four Bernoulli conditions after card 05.

0/5QUESTS

Recall — your gut answer first

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You flip a coin 8 times. Before any formula — how many distinct possible sequences of heads and tails are there in total? And does the probability of getting heads on flip 5 depend in any way on what happened on flips 1–4? Justify in your own words below.

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The two moves for recognising a Bernoulli process

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Every Bernoulli problem rewards two habits: name the trial and the success precisely, then verify all four conditions before applying any formula. Jumping straight to $\binom{n}{k}p^k(1-p)^{n-k}$ without checking the setup is the single most common reason students lose method marks in Statistical Analysis.

The name-and-verify strategy: (1) define one trial and write down the two possible outcomes, (2) state $p$ (probability of success) and confirm it is constant, (3) confirm trials are independent, (4) confirm $n$ is fixed in advance.

Single Bernoulli trial: $P(S) = p$, $P(F) = 1 - p$. Repeated $n$ times under the four conditions $\Rightarrow$ Bernoulli process.

$P(S) = p,\ P(F) = 1-p$

Define "success" first

Write a one-sentence definition of what counts as a success before you write any probability. "Success = getting a six on a die roll." This anchors the rest of the problem and stops you from miscounting later.

Watch for "without replacement"

If items are drawn without replacement (e.g. picking marbles from a bag and not returning them), the probability changes from trial to trial. That breaks the constant-$p$ condition and the setup is NOT Bernoulli.

Fixed $n$, not "until something happens"

"Flip until the first head" is NOT a Bernoulli process because $n$ is not known in advance. Bernoulli requires the number of trials to be set before you start.

What you'll master

Know

Key facts

A Bernoulli trial has exactly two outcomes: success (S) and failure (F)
$P(S) = p$ and $P(F) = 1 - p$ on every trial
The four Bernoulli conditions: two outcomes, fixed $n$, constant $p$, independent trials

Understand

Concepts

Why "without replacement" breaks the constant-$p$ condition
Why "trial until first success" is not Bernoulli (no fixed $n$)
How to reframe a multi-outcome situation as a binary success/failure trial

Can do

Skills

Identify whether a real-world setup satisfies the four Bernoulli conditions
Define the trial and the success outcome precisely
State $n$ and $p$ from the wording of an HSC problem

Key terms

Bernoulli trialA single random experiment with exactly two possible outcomes, conventionally labelled success (S) and failure (F).

Probability of success $p$The probability of the outcome labelled "success" on a single trial. Must be the same value on every trial of a Bernoulli process.

Independent trialsTrials whose outcomes do not influence each other — knowing the result of trial 3 tells you nothing about the result of trial 4.

Bernoulli processA sequence of $n$ Bernoulli trials that satisfies all four conditions: two outcomes, fixed $n$, constant $p$, independent.

With/without replacement"With replacement" preserves constant $p$ (Bernoulli OK). "Without replacement" changes $p$ each trial (not Bernoulli).

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

The four conditions of a Bernoulli process

core concept

A sequence of trials is a Bernoulli process if and only if all four conditions are simultaneously true:

Two outcomes per trial. Each trial results in exactly one of two outcomes, "success" or "failure". (Multi-outcome trials must be reframed as binary first.)
Fixed number of trials $n$. The total count of trials is decided before the experiment begins, not determined by what happens during it.
Constant probability of success $p$. The probability of success has the same value on every trial.
Independent trials. The outcome of one trial does not affect the probability of any other.

Worked through the hook: Roll a fair die 10 times, "success" = rolling a six.

Two outcomes per trial? Yes — either "six" (S) or "not a six" (F). ✓
Fixed $n$? Yes — $n = 10$ is stated in advance. ✓
Constant $p$? Yes — $p = \tfrac{1}{6}$ on every roll because the die is unchanged. ✓
Independent? Yes — rolling the die again is physically independent of previous rolls. ✓

All four conditions hold, so this is a Bernoulli process with $n=10$ and $p = \tfrac{1}{6}$.

Connecting forward. Once a setup satisfies the four Bernoulli conditions, the number of successes $X$ follows a binomial distribution — written $X \sim \mathrm{Bin}(n, p)$. You will meet the binomial PMF in the next lesson. But the binomial formula only makes sense if the underlying process is genuinely Bernoulli.

BINS checklist: (B) Binary outcomes; (I) Independent trials; (N) Number of trials fixed in advance; (S) Same probability $p$ on every trial. All four must hold for the binomial model.

Pause — copy the four Bernoulli conditions (BINS: Binary, Independent, Number fixed, Same $p$) as a four-point checklist into your book.

Quick check: Which of the following situations is NOT a Bernoulli process?

Spotting non-Bernoulli setups

core concept

We just saw that a Bernoulli process requires all four conditions simultaneously: binary outcomes, fixed $n$, constant $p$, and independent trials. That raises a question: what do the three most common non-Bernoulli setups look like, and how do you identify which condition fails in each? This card answers it → sampling without replacement fails independence; variable $n$ fails the "fixed trials" condition; multi-outcome trials fail the "binary" condition.

HSC examiners deliberately include scenarios that fail one Bernoulli condition. Spotting which condition fails is worth as many method marks as solving a binomial calculation. The three most common non-Bernoulli setups are:

Without-replacement sampling. Drawing 5 cards from a deck without putting them back: $p$ of "king" changes after each draw, breaking constant $p$.
Trial until first success. Flipping a coin until the first head: $n$ is random, not fixed in advance. This is a geometric setup, not Bernoulli.
Dependent trials. A weather model where today's rain raises tomorrow's chance of rain: outcomes influence each other, breaking independence.

$$\text{Bernoulli} \iff \text{(2 outcomes) AND (fixed } n\text{) AND (constant } p\text{) AND (independent)}$$

Common mistake. Students assume "two outcomes" alone is enough to call something Bernoulli. It isn't. All four conditions must hold simultaneously. If a question says "draw without replacement", stop — constant $p$ is broken, and the binomial formula is wrong.

Pause — copy the three most common non-Bernoulli setups (without-replacement, variable $n$, multi-outcome) with the specific condition that each violates into your book.

Did you get this? True or false: drawing 4 marbles from a bag containing 7 red and 3 blue marbles WITH replacement, counting reds, is a Bernoulli process.

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

PROBLEM 1 · IDENTIFY THE BERNOULLI PARAMETERS

A student guesses randomly on a 20-question multiple-choice test with four equally likely options per question. (a) Verify this is a Bernoulli process. (b) State $n$ and $p$.

Define: one trial = answering one question. Success = correct answer. Outcomes: correct (S) or incorrect (F). That is exactly two outcomes per trial. ✓

Always name the trial and the success outcome before anything else. This stops you confusing "questions" with "trials" later.

PROBLEM 2 · BERNOULLI OR NOT?

A bag contains 5 red and 5 blue balls. Three balls are drawn without replacement. Let $X$ be the number of red balls drawn. Is this a Bernoulli process? Justify your answer.

Trial = one draw. Success = "red ball". Outcomes per trial: red (S) or blue (F). Two outcomes ✓. Fixed $n = 3$ ✓.

First two conditions look fine — the issue (if any) must be with $p$ or independence.

PROBLEM 3 · REFRAMING A MULTI-OUTCOME SETUP

A fair six-sided die is rolled 12 times. Let $X$ be the number of rolls that show a number greater than 4. Verify this is a Bernoulli process and state $n$ and $p$.

A die has 6 outcomes (1 through 6) — more than two — but we only care about whether the result is greater than 4. Reframe: success = "rolled a 5 or 6", failure = "rolled 1, 2, 3 or 4". Now there are two outcomes per trial ✓.

The two-outcome condition is about how WE classify each trial, not the total number of physical outcomes. Many real-world dichotomies (defective/not, pass/fail, taller than 180/not) are reframings of multi-valued data.

Fill the gap: A fair six-sided die is rolled times and the number of sixes is recorded. This is a Bernoulli process with $n = 15$ and $p = \dfrac{1}{}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Calling "without replacement" a Bernoulli process

If items are drawn from a finite collection and not put back, the probability of success changes from trial to trial. Constant $p$ fails. You CANNOT use the binomial formula. The correct statement is "this is not Bernoulli — constant-$p$ condition fails".

Trap 02

Treating "until first success" as Bernoulli

"Flip a coin until you get a head" is NOT Bernoulli because $n$ is random — it depends on how the coin lands. Bernoulli requires $n$ to be fixed before the experiment starts. This kind of scenario is geometric, not binomial.

Trap 03

Confusing the trial outcome with the count of successes

One trial has TWO outcomes (S or F). The total number of successes $X$ across $n$ trials has many possible values (0, 1, 2, …, $n$). Don't confuse "two outcomes per trial" with "two outcomes for $X$" — the binary structure applies to each individual trial, not to the aggregate count.

Did you get this? True or false: a coin is flipped until the first head appears, and the total number of flips is recorded. This is a Bernoulli process.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A factory tests 50 batteries, each independently having a 2% chance of being defective. State whether this is a Bernoulli process; if so, give $n$ and $p$.

A bag holds 8 red and 4 white marbles. You draw 4 marbles WITHOUT replacement and count the reds. Is this Bernoulli? Justify by checking each condition.

A spinner has 5 equal sectors numbered 1–5. You spin it 25 times and count how many spins land on an odd number. Verify the four Bernoulli conditions and state $n$ and $p$.

A medical test correctly diagnoses a condition 95% of the time. Twelve unrelated patients are tested. Is the count of correct diagnoses a Bernoulli setup? Justify.

A coin is flipped repeatedly UNTIL the third head appears. Is the total number of flips a Bernoulli process? Justify by identifying which condition (if any) fails.

Odd one out: Three of these statements about Bernoulli trials are correct. Which one is NOT?

Revisit your thinking

Earlier you considered rolling a die 10 times and counting sixes. You were asked to identify what counts as a success, the value of $p$, and whether $p$ is the same on every roll.

Success = "rolling a six". $p = \tfrac{1}{6}$ on every roll because the die is unchanged and physical — the result of one roll does not influence the next. All four Bernoulli conditions are satisfied: two outcomes per trial (six / not-six), fixed $n = 10$, constant $p = \tfrac{1}{6}$, independent trials. This means $X$ — the number of sixes — will follow a binomial distribution (next lesson).

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Multiple choice

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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 coin biased so that $P(\text{head}) = 0.6$ is flipped 30 times. State whether this is a Bernoulli process and give $n$ and $p$. (2 marks)

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ApplyBand 43 marks

Q2. A teacher draws 6 cards from a standard 52-card deck without replacement and records the number of aces. State, with full justification by referring to all four Bernoulli conditions, whether the number of aces follows a binomial distribution. (3 marks)

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AnalyseBand 53 marks

Q3. Twenty traffic lights along a route operate independently. Each light is red with probability $0.3$ when a particular driver reaches it. The driver counts how many of the 20 lights are red. Justify why this satisfies the Bernoulli conditions and state $n$, $p$ and $1-p$. Then explain why the count of CONSECUTIVE green lights from the start of the route is NOT a Bernoulli setup. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Two outcomes (defective/not) ✓; $n = 50$ fixed ✓; $p = 0.02$ constant (each battery independent of the rest) ✓; independent ✓. Bernoulli with $n = 50$, $p = 0.02$.

2. Two outcomes (red/white) ✓; $n = 4$ fixed ✓; $p = \tfrac{8}{12}$ on draw 1 but only $\tfrac{7}{11}$ or $\tfrac{8}{11}$ on draw 2 depending on outcome — constant $p$ FAILS. NOT Bernoulli.

3. Reframe: success = odd. $p = \tfrac{3}{5}$ on every spin (1, 3, 5 are odd). Two outcomes ✓; $n = 25$ fixed ✓; constant $p = \tfrac{3}{5}$ ✓; spins independent ✓. Bernoulli with $n = 25$, $p = \tfrac{3}{5}$.

4. Two outcomes (correct/incorrect) ✓; $n = 12$ fixed ✓; $p = 0.95$ constant ✓; patients unrelated $\Rightarrow$ independent ✓. Bernoulli with $n = 12$, $p = 0.95$.

5. $n$ is the total flips needed to get the third head — this is RANDOM, not fixed before the experiment. Fixed-$n$ condition FAILS. NOT Bernoulli (this is a negative binomial setup, beyond syllabus).

Q1 (2 marks): Two outcomes (H or T), $n = 30$ fixed, $p = 0.6$ constant on every flip, flips independent [1]. All four conditions hold $\Rightarrow$ Bernoulli with $n = 30$, $p = 0.6$ [1].

Q2 (3 marks): Two outcomes (ace/not-ace) ✓; $n = 6$ fixed ✓ [1]. $p$ on draw 1 = $\tfrac{4}{52}$, but after one ace is removed $p$ on draw 2 = $\tfrac{3}{51}$ — $p$ is NOT constant [1]. Constant-$p$ condition fails, so not Bernoulli, and the number of aces does NOT follow a binomial distribution [1].

Q3 (3 marks): Part 1: lights operate independently (given), each light is red (S) or not (F), $n = 20$ fixed, $p = 0.3$ constant. All four conditions met ✓ Bernoulli with $n = 20$, $p = 0.3$, $1 - p = 0.7$ [1.5]. Part 2: "consecutive greens from the start" stops as soon as the first red is encountered, so the number of trials is determined by the outcomes — $n$ is NOT fixed in advance, breaking the fixed-$n$ condition [1.5].

Boss battle · The Bernoulli Detective

earn bronze · silver · gold

Five timed questions: name the trial, name the success, check all four conditions, and decide Bernoulli or not. 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 deciding whether each statistical setup is Bernoulli. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

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Module overview · Extension 1 · Checkpoint 1