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Module 10 · L13 of 20 ~35 min ⚡ +90 XP available

Binomial Problems — Single Step

A quality-control engineer tests 10 randomly chosen lightbulbs from a production line where 8% are defective. What is the probability that exactly 2 are defective? Before you can compute anything, you must extract three numbers from the words: the number of trials $n$, the probability of success $p$, and the target count $k$. This lesson trains that translation skill — recognising a binomial scenario and computing one probability cleanly.

Today's hook — A fair coin is flipped 5 times. Before reading on, write down the values of $n$, $p$ and $k$ if you want $P(\text{exactly 3 heads})$, then compute the probability. Compare your answer after card 05.

0/5QUESTS

Recall — your gut answer first

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A spinner lands on red with probability $0.4$. You spin it 6 times and want the probability of exactly 2 reds. Without computing — what are $n$, $p$ and $k$, and which formula will you use?

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The two moves for binomial problems

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Every binomial problem rewards two habits: extract $(n, p, k)$ from the words, then substitute into the formula $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$. The skill is reading the question and translating, not memorising tricks.

The extract-and-substitute strategy: (1) identify the trial (one repeated experiment), (2) read off $n$, $p$, $k$, (3) plug into the formula, evaluate $\binom{n}{k}$, raise to powers, multiply.

$P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$ · $X \sim \text{Bin}(n,p)$

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

Define "success" first

Decide what counts as a success (defective bulb, head, made shot). The value of $p$ is the probability of this success; $1-p$ is the failure probability $q$.

Check the binomial conditions

Fixed $n$ trials, two outcomes per trial, constant $p$, independent trials. If any condition fails, the binomial formula doesn't apply.

Write the formula before computing

Always write $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$ with the actual numbers substituted before evaluating — markers award method marks for this line alone.

What you'll master

Know

Key facts

Binomial formula: $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$
The four binomial conditions (fixed $n$, two outcomes, constant $p$, independence)
The notation $X \sim \text{Bin}(n, p)$ and what each parameter means

Understand

Concepts

Why $\binom{n}{k}$ counts the number of ways to arrange $k$ successes among $n$ trials
How $p^k(1-p)^{n-k}$ gives the probability of one specific sequence of outcomes
Why we identify "success" first — it determines which value is $p$ and which is $1-p$

Can do

Skills

Translate a worded scenario into the values of $n$, $p$, $k$
Substitute correctly into $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$ and evaluate
Recognise and reject scenarios that are not binomial (varying $p$, dependent trials, etc.)

Key terms

Bernoulli trialA single experiment with exactly two outcomes (success/failure) and constant probability $p$ of success. The atomic unit of a binomial process.

Number of trials ($n$)The fixed total number of repeated Bernoulli trials performed in the experiment. Must be known in advance and unchanging.

Probability of success ($p$)The probability of "success" on a single trial. Constant across all trials; $0 \leq p \leq 1$. Failure probability is $q = 1-p$.

Target count ($k$)The number of successes you want the probability for. Always satisfies $0 \leq k \leq n$.

Binomial coefficient $\binom{n}{k}$The number of ways to choose which $k$ of the $n$ trials are successes; equals $\dfrac{n!}{k!(n-k)!}$.

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

Extracting $n$, $p$, $k$ from a worded problem

core concept

The single biggest skill in binomial problems is reading the question and answering three questions:

$n$ = ? How many independent trials are performed? (Look for "10 bulbs", "5 flips", "20 attempts".)
$p$ = ? What is the probability of success on one trial? (Define success first — defective, head, made shot.)
$k$ = ? How many successes are we asking about? (Look for "exactly 2", "exactly 3".)

Worked through the hook: A fair coin flipped 5 times, find $P(\text{exactly 3 heads})$:

$n = 5$ (5 flips, fixed).
Success = head; $p = 0.5$, so $1-p = 0.5$.
$k = 3$ (exactly 3 heads).
$P(X=3) = \displaystyle\binom{5}{3}(0.5)^3(0.5)^2 = 10 \times 0.125 \times 0.25 = 10 \times 0.03125 = 0.3125 = \dfrac{5}{16}$.

Connecting to the formula structure. $\binom{5}{3} = 10$ counts the 10 distinct arrangements (HHHTT, HHTHT, …); $(0.5)^3(0.5)^2$ is the probability of any one such arrangement. Multiplying gives the total probability of "any way of getting exactly 3 heads".

The single biggest skill in binomial problems is reading the question and answering three questions:

Pause — copy the three-question extraction method: $n$ (trials), $p$ (success probability), $k$ (successes asked), with the hook example worked through into your book.

Quick check: A multiple-choice quiz has 8 questions, each with 4 options. A student guesses every question. For $P(\text{exactly 3 correct})$, what are $n$, $p$, $k$?

Substituting and evaluating cleanly

core concept

We just saw that reading a worded problem requires answering three explicit questions: what is $n$ (number of trials), $p$ (success probability), and $k$ (number of successes asked about). That raises a question: once you have $(n,p,k)$, how do you evaluate $\binom{n}{k}p^k(1-p)^{n-k}$ cleanly in three steps without calculator errors? This card answers it → compute $\binom{n}{k}$, $p^k$, and $(1-p)^{n-k}$ separately, then multiply.

Once you have $(n, p, k)$, the calculation has three pieces that must each be computed and then multiplied:

$\binom{n}{k}$: use the formula $\dfrac{n!}{k!(n-k)!}$ or your calculator's $\text{nCr}$ button.
$p^k$: the success probability raised to the success count.
$(1-p)^{n-k}$: the failure probability raised to the failure count.

Example: A spinner lands on red with $p = 0.4$. Spun 6 times, find $P(\text{exactly 2 reds})$.

$n = 6$, $p = 0.4$, $k = 2$.
$\binom{6}{2} = 15$.
$p^k = 0.4^2 = 0.16$.
$(1-p)^{n-k} = 0.6^4 = 0.1296$.
$P(X=2) = 15 \times 0.16 \times 0.1296 \approx 0.3110$.

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

Common mistake. Students sometimes raise $p$ to the wrong power, swapping $k$ and $n-k$. Remember: $p$ goes with $k$ (the success count) and $(1-p)$ goes with $n-k$ (the failure count).

Once you have $(n, p, k)$, the calculation has three pieces that must each be computed and then multiplied:

Pause — copy the three-part evaluation: compute $\binom{n}{k}$, then $p^k$, then $(1-p)^{n-k}$, then multiply into your book.

Did you get this? True or false: if $X \sim \text{Bin}(10, 0.3)$, then $P(X = 4) = \binom{10}{4}(0.3)^4(0.7)^6$.

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

PROBLEM 1 · QUALITY CONTROL

A factory's lightbulb production line has a defect rate of $8\%$. A quality inspector selects $10$ bulbs at random. Find $P(\text{exactly 2 are defective})$.

Extract: success = "defective bulb". $n = 10$, $p = 0.08$, $k = 2$. So $X \sim \text{Bin}(10, 0.08)$.

Define "success" as the event you're counting. Here we count defective bulbs, so $p$ is the defect rate $= 0.08$.

PROBLEM 2 · BASKETBALL FREE THROWS

A player makes $70\%$ of her free throws. In a practice set she takes $12$ shots. Find $P(\text{she makes exactly 9})$.

Extract: success = "made shot". $n = 12$, $p = 0.7$, $k = 9$. So $X \sim \text{Bin}(12, 0.7)$.

12 fixed shots, two outcomes (made/missed), constant probability $0.7$, independent trials — all four binomial conditions are met.

PROBLEM 3 · MULTIPLE-CHOICE GUESSING

A 6-question multiple-choice quiz gives 5 options per question. A student guesses every question. Find $P(\text{exactly 4 correct})$.

Extract: success = "correct answer". $n = 6$, $p = \tfrac{1}{5} = 0.2$, $k = 4$. So $X \sim \text{Bin}(6, 0.2)$.

Each question has 5 options, only 1 correct, so $p = 1/5$. Trials are independent because the student is guessing each time.

Fill the gap: A coin is flipped 4 times. $P(\text{exactly 2 heads}) = \binom{4}{2}(0.5)^2(0.5)^2 = \dfrac{}{16}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Swapping $k$ and $n-k$ in the exponents

Students sometimes write $\binom{n}{k}p^{n-k}(1-p)^k$. The success probability $p$ must be raised to the SUCCESS count $k$, and the failure probability $(1-p)$ to the FAILURE count $n-k$. Mixing these gives a wrong probability with no method marks.

Trap 02

Forgetting to define "success"

If the question asks about defective bulbs and you treat "non-defective" as success, you set $p$ wrong. Always state clearly which event is success — and check that $p$ matches it.

Trap 03

Applying the formula when conditions fail

"Drawing 3 balls from a bag without replacement" is NOT binomial — $p$ changes after each draw. Binomial requires fixed $n$, constant $p$, two outcomes per trial, and independent trials. If any condition fails, the formula does not apply.

Did you get this? True or false: drawing 5 cards from a standard deck without replacement and counting hearts can be modelled by a binomial distribution.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A fair die is rolled $5$ times. Find $P(\text{exactly 2 sixes})$. State $n$, $p$, $k$, write the formula, then evaluate.

In a population, $30\%$ of people are left-handed. A sample of $8$ people is taken. Find $P(\text{exactly 3 are left-handed})$.

A multiple-choice quiz has $10$ questions, each with $4$ options. A student guesses every question. Find $P(\text{exactly 5 correct})$.

A basketball player makes $80\%$ of her free throws. She takes $7$ shots. Find $P(\text{she makes exactly 6})$.

A surgical procedure has a $95\%$ success rate. Out of $20$ patients, find $P(\text{exactly 18 successful})$.

Odd one out: Three of these scenarios CAN be modelled by a binomial distribution. Which one CANNOT?

Revisit your thinking

Earlier you wrote down $n$, $p$, $k$ for the spinner problem (red with $p=0.4$, spun 6 times, exactly 2 reds).

The values are $n=6$, $p=0.4$, $k=2$. Substituting: $P(X=2) = \binom{6}{2}(0.4)^2(0.6)^4 = 15 \times 0.16 \times 0.1296 \approx 0.311$. The discipline of writing the formula with substituted numbers before evaluating is what earns the method marks.

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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 die is rolled $4$ times. Find $P(\text{exactly 1 six})$. (2 marks)

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ApplyBand 43 marks

Q2. A vaccine is effective in $85\%$ of patients. In a sample of $15$ patients, find $P(\text{exactly 13 respond})$. Round to 4 decimal places. (3 marks)

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AnalyseBand 53 marks

Q3. $X \sim \text{Bin}(n, p)$. Given $P(X = 0) = 0.1296$ and $p = 0.4$, find $n$, then compute $P(X = 2)$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $n=5, p=1/6, k=2$. $P = \binom{5}{2}(1/6)^2(5/6)^3 = 10 \cdot \frac{1}{36} \cdot \frac{125}{216} = \frac{1250}{7776} \approx 0.1608$.

2. $n=8, p=0.3, k=3$. $P = \binom{8}{3}(0.3)^3(0.7)^5 = 56 \cdot 0.027 \cdot 0.16807 \approx 0.2541$.

3. $n=10, p=0.25, k=5$. $P = \binom{10}{5}(0.25)^5(0.75)^5 = 252 \cdot 0.0009766 \cdot 0.2373 \approx 0.0584$.

4. $n=7, p=0.8, k=6$. $P = \binom{7}{6}(0.8)^6(0.2)^1 = 7 \cdot 0.2621 \cdot 0.2 \approx 0.3670$.

5. $n=20, p=0.95, k=18$. $P = \binom{20}{18}(0.95)^{18}(0.05)^2 = 190 \cdot 0.3972 \cdot 0.0025 \approx 0.1887$.

Q1 (2 marks): $n=4, p=1/6, k=1$ [1]. $P = \binom{4}{1}(1/6)^1(5/6)^3 = 4 \cdot \frac{1}{6} \cdot \frac{125}{216} = \frac{500}{1296} \approx 0.3858$ [1].

Q2 (3 marks): $n=15, p=0.85, k=13$ [1]. $P = \binom{15}{13}(0.85)^{13}(0.15)^2$ [1] $= 105 \cdot 0.1209 \cdot 0.0225 \approx 0.2856$ [1].

Q3 (3 marks): $P(X=0) = (0.6)^n = 0.1296$ [1]. Since $0.6^4 = 0.1296$, $n = 4$ [1]. $P(X=2) = \binom{4}{2}(0.4)^2(0.6)^2 = 6 \cdot 0.16 \cdot 0.36 = 0.3456$ [1].

Boss battle · The Binomial Single-Shot

earn bronze · silver · gold

Five timed questions on extracting $n$, $p$, $k$ and computing $P(X = k)$. 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 single-step binomial questions. 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 3