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Module 10 · L15 of 20 ~40 min ⚡ +90 XP available

Binomial Problems — Contextual

A quality-control engineer pulls 20 chips off the line knowing 3% are defective. A geneticist crosses pea plants where 75% should yield round seeds. A netball goal shooter sinks 80% from the post. Three very different contexts — one model. This lesson trains you to translate a real-world story into the binomial parameters $(n, p, k)$ and pick the right tail of the distribution.

Today's hook — A basketball player has a 60% free-throw success rate. She shoots 10 free throws in a game. Before reading on, write down (a) what $n$, $p$ and a "success" should be, and (b) the expression for "exactly 7 baskets" using the binomial formula. Compare your set-up after card 05.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

A coin is biased so that $P(\text{heads}) = 0.7$. It is tossed 5 times. Before using any formula — identify $n$, $p$, and "success", then write (don't evaluate) the expression for $P(\text{exactly 3 heads})$. Sketch your reasoning below.

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The two moves for contextual problems

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Every contextual binomial problem rewards two habits: translate the story into $(n, p)$ by naming a clear "success", then read the wording carefully to choose the right tail — exactly, at least, at most, or between. The wording controls the calculation.

The translate-and-tail strategy: (1) name the trial and the success event, (2) extract $n$ (number of trials) and $p$ (success probability), (3) parse the wording — "exactly $k$" is $P(X=k)$, "at least $k$" is $P(X\geq k)$, "at most $k$" is $P(X\leq k)$.

$P(X = k) = \dbinom{n}{k} p^k (1-p)^{n-k}$, $X \sim B(n, p)$

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

Name the success

Write the success event explicitly: "success = chip is defective" or "success = seed is round". The probability $p$ is the probability of that exact event on one trial.

"At least one" trick

$P(X \geq 1) = 1 - P(X = 0)$. The complement is much faster than summing $P(X=1) + P(X=2) + \ldots + P(X=n)$.

Check the assumptions

Binomial requires fixed $n$, two outcomes per trial, constant $p$, and independent trials. Sampling without replacement from a small population breaks independence.

What you'll master

Know

Key facts

Binomial PMF: $P(X = k) = \dbinom{n}{k} p^k (1-p)^{n-k}$
Conditions: fixed $n$, two outcomes, constant $p$, independent trials
Complement rule: $P(X \geq 1) = 1 - P(X = 0)$

Understand

Concepts

How to translate a real-world story into $(n, p, k)$ parameters
Why naming the "success" event determines the value of $p$
How the wording ("exactly", "at least", "at most") maps to the correct tail

Can do

Skills

Solve binomial problems in quality control, genetics, sport and survey contexts
Use the complement rule to evaluate "at least one" probabilities efficiently
Identify when the binomial model is — and is not — appropriate

Key terms

Bernoulli trialA single experiment with two outcomes ("success" with probability $p$, "failure" with probability $1-p$).

Binomial random variable$X \sim B(n, p)$ counts the number of successes in $n$ independent Bernoulli trials with constant $p$.

SuccessThe outcome being counted in the problem — must be defined explicitly. $p$ is the probability of this specific outcome.

Tail probabilityA cumulative probability such as $P(X \leq k)$ or $P(X \geq k)$ obtained by summing PMF values across the relevant range.

ComplementFor any event $A$, $P(A) = 1 - P(A^c)$. Used to convert "at least one" into a single term: $1 - P(X = 0)$.

ME12-5NESA outcome: applies statistical processes to evaluate inferences and conclusions, including the binomial distribution.

Translating a story into $(n, p, k)$

core concept

Contextual problems test whether you can extract the right model from a paragraph of English. The four-step strategy never fails:

Name the trial. What is the repeated experiment? (Inspecting a chip, crossing a plant, taking a shot.)
Name the success. Which outcome is being counted? Write it explicitly.
Extract $(n, p)$. $n$ = number of trials; $p$ = $P(\text{success})$ on one trial.
Read the tail. "Exactly $k$" = $P(X=k)$; "at least $k$" = $P(X \geq k)$; "at most $k$" = $P(X \leq k)$; "more than $k$" = $P(X > k)$.

Worked through the hook: Basketball player with 60% free-throw rate, 10 shots, exactly 7 baskets.

Trial: one free throw. Success: making the basket.
$n = 10$, $p = 0.6$, $k = 7$, so $X \sim B(10, 0.6)$.
$P(X = 7) = \dbinom{10}{7}(0.6)^7(0.4)^3 = 120 \cdot 0.0279936 \cdot 0.064 \approx 0.2150$.
Sanity check: $E(X) = np = 6$, so $k=7$ is just above the mean — a probability around 0.2 is reasonable.

The wording table. "Exactly $k$" $\to P(X=k)$. "At least $k$" $\to P(X \geq k) = 1 - P(X \leq k-1)$. "At most $k$" $\to P(X \leq k)$. "Fewer than $k$" $\to P(X \leq k-1)$. "More than $k$" $\to P(X \geq k+1)$. Misreading the wording is the single biggest source of lost marks.

Contextual problems test whether you can extract the right model from a paragraph of English. The four-step strategy never fails:

Pause — copy the four-step English-to-binomial translation method with the pea-plant example ($n$, $p$, success defined) into your book.

Quick check: A factory produces light bulbs and 4% are defective. A sample of 12 bulbs is tested. What expression gives the probability that exactly 2 bulbs are defective?

The "at least one" shortcut

core concept

We just saw the four-step English-to-binomial translation: name the trial, define success, extract $n$ and $p$, read the tail. That raises a question: the phrase "at least one" appears in almost every contextual binomial problem — why is the complement $1-(1-p)^n$ always faster than summing $P(X=1)+P(X=2)+\cdots+P(X=n)$? This card answers it → direct summation requires $n$ terms; the complement reduces it to one calculation.

"At least one defective" is the most common contextual phrase. Computing $P(X \geq 1)$ directly means summing $n$ terms, which is wasteful. The complement gives a single calculation:

$P(X \geq 1) = 1 - P(X = 0) = 1 - (1-p)^n$.

Example: A geneticist crosses pea plants where 25% of offspring should be wrinkled (recessive). Out of 8 offspring, what is the probability that at least one is wrinkled?

Trial: one offspring. Success: wrinkled. $n=8$, $p=0.25$.
$P(X \geq 1) = 1 - P(X = 0) = 1 - (0.75)^8 \approx 1 - 0.1001 = 0.8999$.

$$P(X \geq 1) = 1 - (1-p)^n, \qquad P(X \leq k) = \sum_{i=0}^{k}\binom{n}{i}p^i(1-p)^{n-i}$$

Common mistake. Students often write $P(X \geq 1) = P(X = 1)$, which only counts one term. "At least one" includes 1, 2, 3, … all the way up to $n$. The complement $1 - P(X=0)$ captures all of those in one step.

"At least one defective" is the most common contextual phrase. Computing $P(X \geq 1)$ directly means summing $n$ terms, which is wasteful. The complement gives a single calculation:

Pause — copy the "at least one" shortcut: $P(X\geq1)=1-P(X=0)=1-(1-p)^n$ and verify it is faster than summing $n$ individual terms into your book.

Did you get this? True or false: if $X \sim B(20, 0.05)$ represents the number of defective items in a sample of 20, then $P(X \geq 1) = 1 - (0.95)^{20}$.

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

PROBLEM 1 · QUALITY CONTROL

A factory's chip line has a 3% defect rate. A quality inspector samples 20 chips. Find the probability that (a) exactly 1 chip is defective, (b) at least one chip is defective.

Trial: inspect one chip. Success: chip is defective. $n = 20$, $p = 0.03$, so $X \sim B(20, 0.03)$.

Always define the trial and "success" before reaching for the formula. Here we count defective chips, so $p$ is the defect rate.

PROBLEM 2 · BIOLOGY / GENETICS

In a Mendelian cross, 75% of pea offspring should yield round seeds (dominant), 25% wrinkled. A geneticist examines 6 seeds. Find the probability that more than 4 are round.

Trial: inspect one seed. Success: round. $n = 6$, $p = 0.75$, $X \sim B(6, 0.75)$. "More than 4" means $X = 5$ or $X = 6$.

"More than 4" excludes 4 itself, so the relevant values are 5 and 6 — we add those two probabilities.

PROBLEM 3 · SURVEY / SPORT

A survey finds 40% of high-school students play organised sport. A class of 15 is randomly selected. Find the probability that at most 5 students play organised sport.

Trial: pick one student. Success: plays organised sport. $n = 15$, $p = 0.4$, $X \sim B(15, 0.4)$. "At most 5" means $X \leq 5$, i.e. $X \in \{0, 1, 2, 3, 4, 5\}$.

"At most $k$" includes $k$ itself and everything below it. Six terms to sum here.

Fill the gap: A netball shooter sinks 80% of attempts. In 5 attempts the probability she misses all 5 is $(0.)^5 \approx 0.00032$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Mis-naming the success

If a problem states "4% of bulbs are defective" and asks for the probability that none are defective, students sometimes set $p = 0.96$ (the success rate of "good bulb") but then write $P(X = 0)$ where $X$ counts defectives. You must commit to one definition of success and stay with it throughout the calculation.

Trap 02

"At least one" $\neq$ "exactly one"

$P(X \geq 1)$ includes $X = 1, 2, 3, \ldots, n$ — not just $X = 1$. Always use the complement: $P(X \geq 1) = 1 - P(X = 0)$. Writing only $P(X = 1)$ undercounts and is a common source of full-mark loss.

Trap 03

Independence assumption ignored

Binomial requires independent trials with constant $p$. Sampling without replacement from a small population (e.g., 5 cards from a deck of 20) breaks independence — $p$ changes after each draw. Always check whether the trials really are independent before applying the binomial model.

Did you get this? True or false: drawing 5 cards from a standard 52-card deck without replacement and counting hearts is well-modelled by a binomial distribution.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A coin is biased with $P(\text{heads}) = 0.7$. It is tossed 10 times. Find $P(X = 6)$ where $X$ counts heads.

A vaccine is effective in 90% of cases. 8 patients are given the vaccine. Find the probability that all 8 are protected.

A multiple-choice quiz has 10 questions, each with 4 options. A student guesses every answer. Find the probability of getting at least 3 correct.

In a country town, 35% of households own a pet. A researcher randomly samples 12 households. Find the probability that fewer than 4 own a pet.

A goalkeeper saves 25% of penalty kicks. Find the probability that out of 5 penalties she saves at least one. Then identify why the binomial model is appropriate here.

Odd one out: Three of these contexts are valid binomial settings. Which one is NOT?

Revisit your thinking

Earlier you set up "biased coin, 5 tosses, exactly 3 heads" and identified $n$, $p$ and "success".

The translation is: trial = one toss, success = heads, $n=5$, $p=0.7$, $k=3$. So $P(X=3) = \binom{5}{3}(0.7)^3(0.3)^2 = 10 \cdot 0.343 \cdot 0.09 = 0.3087$. The most common error is reversing the exponents — $p^k(1-p)^{n-k}$, not $p^{n-k}(1-p)^k$.

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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 box contains a large number of light bulbs, 5% of which are defective. A sample of 10 bulbs is taken. Find the probability that exactly 2 bulbs are defective. (2 marks)

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ApplyBand 43 marks

Q2. A tennis player wins 65% of her service games. In a match she serves 8 service games. Find the probability that she wins at least 6 of them. (3 marks)

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AnalyseBand 53 marks

Q3. A pharmaceutical company claims a treatment is effective in 80% of cases. A trial is run on 12 patients. Find the probability that more than 9 patients are successfully treated, and comment on whether the result would still be unusual if only 7 patients had succeeded. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $P(X=6) = \binom{10}{6}(0.7)^6(0.3)^4 = 210 \cdot 0.117649 \cdot 0.0081 \approx 0.2001$.

2. $P(X=8) = (0.9)^8 \approx 0.4305$.

3. $X \sim B(10, 0.25)$. $P(X=0)=(0.75)^{10}\approx 0.0563$; $P(X=1)=10(0.25)(0.75)^9\approx 0.1877$; $P(X=2)=45(0.25)^2(0.75)^8\approx 0.2816$. $P(X\geq 3) = 1 - (0.0563+0.1877+0.2816) \approx 0.4744$.

4. $X \sim B(12, 0.35)$. $P(X=0)\approx 0.00569$; $P(X=1)\approx 0.03680$; $P(X=2)\approx 0.10911$; $P(X=3)\approx 0.19580$. $P(X<4) = P(X\leq 3) \approx 0.3474$.

5. $P(X \geq 1) = 1 - (0.75)^5 = 1 - 0.2373 \approx 0.7627$. Binomial valid: fixed $n = 5$, two outcomes (save / no save), constant $p$ per kick, kicks independent.

Q1 (2 marks): $X \sim B(10, 0.05)$ [1]. $P(X=2) = \binom{10}{2}(0.05)^2(0.95)^8 = 45 \cdot 0.0025 \cdot 0.6634 \approx 0.0746$ [1].

Q2 (3 marks): $X \sim B(8, 0.65)$ [1]. $P(X=6) = \binom{8}{6}(0.65)^6(0.35)^2 \approx 0.2587$; $P(X=7)\approx 0.1373$; $P(X=8)\approx 0.0319$ [1]. $P(X\geq 6) \approx 0.4279$ [1].

Q3 (3 marks): $X \sim B(12, 0.8)$ [1]. $P(X=10)\approx 0.2835$; $P(X=11)\approx 0.2062$; $P(X=12)\approx 0.0687$; so $P(X>9)\approx 0.5584$ [1]. $E(X)=9.6$, so 7 successes is well below the mean; computing $P(X\leq 7)\approx 0.0726$ shows it is unusual (less than 10% chance) [1].

Boss battle · The Binomial Translator

earn bronze · silver · gold

Five timed questions translating real-world contexts into 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 contextual binomial questions. Lighter alternative to the boss.

Mark lesson as complete

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Module overview · Extension 1 · Checkpoint 4