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Module 10 · L04 of 20 ~40 min ⚡ +90 XP available

The Binomial Random Variable

A pharmaceutical company tests a new vaccine on 50 volunteers, each with an 80% chance of developing immunity. How likely is it that exactly 42 develop immunity? A genetics researcher predicts that 1 in 4 offspring will show a recessive trait — in 16 offspring, what is the chance that exactly 5 show it? These questions all have one universal answer: the binomial PMF, $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$.

Today's hook — A coin biased so that $P(\text{H}) = 0.7$ is flipped 5 times. Before reading on, write down how many sequences of 5 flips contain exactly 3 heads, and explain why $P(X = 3)$ is NOT simply $0.7^3$. Compare your answer after card 05.

0/5QUESTS

Recall — your gut answer first

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A fair coin is flipped 4 times. Before any formula — how many distinct sequences of 4 flips give exactly 2 heads? List a few of them. Why is this counting question essential before you can talk about $P(X = 2)$?

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The two moves for binomial probability

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Every binomial problem rewards two habits: identify $n$, $p$, $k$ exactly, then apply the PMF in three pieces (the combination, the success probability raised to $k$, and the failure probability raised to $n-k$). Skipping either step is the most common cause of arithmetic and method errors.

The extract-and-apply strategy: (1) read the wording carefully and write down $n$ (trials), $p$ (success probability), $k$ (target number of successes), (2) write the PMF in three labelled pieces — ways, successes, failures — (3) substitute and simplify carefully.

$P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$ · $X \sim \mathrm{Bin}(n,p)$, $k \in \{0,1,\dots,n\}$

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

Verify the four conditions first

Before reaching for the formula, check that $n$ is fixed, $p$ is constant, the trials are independent, and each has two outcomes. Without these, $X$ is not binomial and the PMF gives a wrong answer.

Exponents must add to $n$

In $p^k(1-p)^{n-k}$, the two exponents sum to $n$. If your exponents don't add to the total trials, you've miscounted — restart by writing $n$, $p$, $k$ first.

$\binom{n}{k}$ is the order count

The coefficient $\binom{n}{k}$ counts the number of distinct orderings of $k$ successes among $n$ trials. Without it you'd only get the probability of ONE specific sequence, not all sequences with $k$ successes.

What you'll master

Know

Key facts

Binomial notation: $X \sim \mathrm{Bin}(n,p)$ where $X$ is the count of successes
The binomial PMF: $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$ for $k = 0, 1, \ldots, n$
$\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$ counts the orderings of $k$ successes in $n$ trials

Understand

Concepts

Why each piece of the PMF (combination, $p^k$, $(1-p)^{n-k}$) is needed
Why the binomial PMF only applies when the four Bernoulli conditions hold
How the PMF probabilities sum to 1 across $k = 0, 1, \ldots, n$

Can do

Skills

Identify $n$, $p$, $k$ from the wording of an HSC problem
Compute $P(X = k)$ using the binomial PMF, with exact or decimal answers
Recognise when a setup is suitable for the binomial PMF and when it is not

Key terms

Binomial random variable$X$ = number of successes in $n$ Bernoulli trials, written $X \sim \mathrm{Bin}(n, p)$. $X$ can take any integer value from 0 to $n$.

Probability mass function (PMF)A formula giving $P(X = k)$ for each possible value $k$ of a discrete random variable. For binomial: $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$.

Binomial coefficient $\binom{n}{k}$$\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$ — the number of ways to choose $k$ items from $n$, or equivalently the number of orderings of $k$ successes among $n$ trials.

$n$ and $p$ (parameters)The two parameters that fully specify a binomial distribution: $n$ is the number of trials, $p$ is the per-trial success probability.

Failure exponent $n-k$The number of failures is $n - k$ — whatever is left over after $k$ successes out of $n$ trials. The exponent of $(1-p)$ in the PMF.

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

The binomial PMF, piece by piece

core concept

Given a Bernoulli process with parameters $n$ and $p$, the number of successes $X$ has the probability mass function:

$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \qquad k = 0, 1, 2, \ldots, n.$$

The formula has three pieces, each with a clear meaning:

$\binom{n}{k}$ — the number of orderings of $k$ successes among $n$ trials (e.g. for "3 heads in 5 flips", $\binom{5}{3} = 10$ sequences).
$p^k$ — the probability that those $k$ chosen positions are all successes.
$(1-p)^{n-k}$ — the probability that the remaining $n-k$ positions are all failures.

Worked through the hook: Coin with $P(\text{H}) = 0.7$, flipped 5 times, find $P(X=3)$.

$n = 5$, $p = 0.7$, $k = 3$, so $n - k = 2$.
$\binom{5}{3} = \dfrac{5!}{3!\,2!} = 10$ — there are 10 different orderings of 3 heads in 5 flips.
$P(X = 3) = 10 \cdot (0.7)^3 \cdot (0.3)^2 = 10 \cdot 0.343 \cdot 0.09 = 0.3087$.

So roughly a 30.87% chance of exactly 3 heads — not just $0.7^3 = 0.343$, because we have to count all 10 orderings AND multiply by the failure probability for the other 2 flips.

Connecting back. The binomial PMF is only valid when the four Bernoulli conditions (L03) all hold. If even one fails (e.g. drawing without replacement), the PMF is the wrong model — you'd need a different distribution.

Binomial PMF: $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$ for $k=0,1,\ldots,n$. Three factors: $\binom{n}{k}$ = number of arrangements; $p^k$ = probability of $k$ successes; $(1-p)^{n-k}$ = probability of $n-k$ failures.

Pause — copy the binomial PMF $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$ and explain in one sentence what each of the three factors represents into your book.

Quick check: If $X \sim \mathrm{Bin}(8, 0.4)$, which expression correctly gives $P(X = 3)$?

When (and when not) to use the binomial PMF

core concept

We just saw that $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$ where $\binom{n}{k}$ counts arrangements, $p^k$ is the probability of $k$ successes, and $(1-p)^{n-k}$ the probability of $n-k$ failures. That raises a question: after deriving the formula, how do you verify the four Bernoulli conditions before using it, and what are the common setups where it does not apply? This card answers it → always run BINS first: without-replacement sampling, variable $n$, and unequal $p$ each invalidate the model.

The PMF is a powerful tool, but only inside the domain where it is valid. Always run the four-condition check before substituting.

Use $\mathrm{Bin}(n,p)$ when: $n$ fixed in advance, two outcomes per trial, constant $p$, trials independent.
Do NOT use when: trials are dependent, $p$ varies, $n$ is determined by the experiment, or there are more than two effective outcomes per trial without a meaningful dichotomy.

Quick contrast. Compare two scenarios:

"A bag contains 50 marbles, 20 red. You draw 6 marbles WITH replacement. Find $P(\text{exactly 2 red})$." ⇒ Bernoulli holds. $X \sim \mathrm{Bin}(6, 0.4)$. $P(X=2) = \binom{6}{2}(0.4)^2(0.6)^4 = 15 \cdot 0.16 \cdot 0.1296 = 0.31104$. ✓
"Same setup but WITHOUT replacement." ⇒ $p$ changes. Bernoulli fails. The binomial PMF is wrong here.

$$\sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} = (p + (1-p))^n = 1$$

Common mistake. Computing $P(X = k)$ when the underlying process is not Bernoulli. The formula will give you a number, but that number is meaningless — it does not represent the true probability. Always verify the four conditions before substituting.

The PMF is a powerful tool, but only inside the domain where it is valid. Always run the four-condition check before substituting.

Pause — copy the four-condition validity checklist (BINS) and three common disqualifying setups (without replacement, variable $n$, non-constant $p$) into your book.

Did you get this? True or false: for $X \sim \mathrm{Bin}(10, 0.3)$, the exponents in $P(X=4)$ are $p^4$ and $(1-p)^6$, summing to 10.

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

PROBLEM 1 · STRAIGHT APPLICATION OF THE PMF

A biased coin lands heads with probability $0.4$. It is flipped 6 times. Let $X$ be the number of heads. Find $P(X = 2)$, giving an exact answer.

Bernoulli check: two outcomes per flip ✓, $n = 6$ fixed ✓, $p = 0.4$ constant ✓, flips independent ✓. So $X \sim \mathrm{Bin}(6, 0.4)$. Identify $k = 2$, $n - k = 4$.

Always verify the four conditions first, then list $n$, $p$, $k$ explicitly. Skipping this is the most common error.

PROBLEM 2 · "AT LEAST" CALCULATIONS

A multiple-choice test has 10 questions, each with 4 equally likely options. A student guesses every answer. Find the probability that the student gets at least 1 question correct.

$X \sim \mathrm{Bin}(10, 0.25)$ (Bernoulli conditions hold for random independent guessing). We want $P(X \geq 1)$.

"At least one" calculations are much faster via the complement: $P(X \geq 1) = 1 - P(X = 0)$.

PROBLEM 3 · GENETICS APPLICATION

A genetic trait appears in offspring with probability $\tfrac{1}{4}$. A family has 8 children, considered independent. Find the probability that exactly 3 children show the trait.

Bernoulli check: each child either shows the trait (S) or does not (F) ✓; $n = 8$ fixed ✓; $p = \tfrac{1}{4}$ on each child ✓; children independent (assumed) ✓. So $X \sim \mathrm{Bin}(8, \tfrac{1}{4})$. We want $P(X = 3)$.

Independence is stated in the question (a modelling assumption). In reality siblings may share environmental factors, but for HSC we accept the stated independence.

Fill the gap: If $X \sim \mathrm{Bin}(7, 0.5)$, then $P(X = 4) = \binom{7}{}(0.5)^{}(0.5)^{}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the binomial coefficient $\binom{n}{k}$

Writing $p^k(1-p)^{n-k}$ alone gives only the probability of ONE specific sequence (e.g. SSSFF). The coefficient $\binom{n}{k}$ counts how many such sequences exist. Without it your answer is too small by a factor of $\binom{n}{k}$.

Trap 02

Swapping the exponents on $p$ and $(1-p)$

The success probability $p$ is raised to the number of successes $k$, not the number of failures. Writing $p^{n-k}(1-p)^k$ gives the probability for the WRONG count of successes. Sanity check: $p^k$ should appear with the number you're asking about ("$X = k$").

Trap 03

Applying the formula when the conditions fail

Plugging into $\binom{n}{k}p^k(1-p)^{n-k}$ for a setup that isn't Bernoulli (e.g. sampling without replacement, or trials that depend on each other) produces a wrong probability. Verify the four conditions BEFORE substituting — this single check separates Band 5 from Band 4 work.

Did you get this? True or false: if $X \sim \mathrm{Bin}(10, 0.2)$, then $P(X = 0) = (0.8)^{10}$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Let $X \sim \mathrm{Bin}(5, 0.3)$. Find $P(X = 2)$, leaving your answer correct to 4 decimal places.

A fair die is rolled 12 times. Find the probability of getting exactly 4 sixes. Use $X \sim \mathrm{Bin}(12, \tfrac{1}{6})$.

A salesman makes calls; each call results in a sale with probability $0.2$, independent of other calls. He makes 10 calls. Find $P(X \geq 1)$ using the complement.

$X \sim \mathrm{Bin}(4, 0.5)$. Verify that $\displaystyle\sum_{k=0}^4 P(X=k) = 1$ by computing each probability and summing.

A vaccine has an 85% success rate (immunity per dose, independent). It is given to 20 people. Write down (do not evaluate) the expression for $P(\text{exactly 17 develop immunity})$, clearly stating $n$, $p$ and $k$.

Odd one out: Three of these statements about $X \sim \mathrm{Bin}(n, p)$ are correct. Which one is NOT?

Revisit your thinking

Earlier you were asked how many sequences of 5 flips contain exactly 3 heads, and why $P(X = 3)$ is not just $0.7^3$.

The answer to the count is $\binom{5}{3} = 10$ — that is, ten distinct orderings such as HHHTT, HHTHT, HTHTH, … The probability $P(X = 3)$ is NOT just $0.7^3$ because $0.7^3$ alone is the probability of just ONE such sequence (say HHHTT). We must multiply by the number of orderings AND by $(0.3)^2$ for the two tails. So $P(X=3) = 10 \cdot (0.7)^3 \cdot (0.3)^2 = 0.3087$. The three pieces of the PMF correspond to: count of orderings, success probabilities, failure probabilities.

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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. Let $X \sim \mathrm{Bin}(6, 0.5)$. Find $P(X = 4)$ as an exact fraction. (2 marks)

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ApplyBand 43 marks

Q2. A multiple-choice test has 15 questions, each with 4 options. A student guesses randomly. Find the probability that the student gets exactly 5 questions correct. Give the answer correct to 4 decimal places. (3 marks)

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AnalyseBand 53 marks

Q3. A pharmaceutical trial gives a drug to 12 patients independently. Each patient has a $0.7$ probability of responding favourably. Find the probability that at least 10 patients respond favourably. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $P(X = 2) = \binom{5}{2}(0.3)^2(0.7)^3 = 10 \cdot 0.09 \cdot 0.343 = 0.3087$.

2. $P(X = 4) = \binom{12}{4}\left(\tfrac{1}{6}\right)^4\left(\tfrac{5}{6}\right)^8 = 495 \cdot \tfrac{1}{1296} \cdot \tfrac{390625}{1679616} \approx 0.0888$.

3. $P(X \geq 1) = 1 - (0.8)^{10} = 1 - 0.1074 \approx 0.8926$.

4. $\binom{4}{k}$ for $k = 0,1,2,3,4$ are $1, 4, 6, 4, 1$. With $p = 0.5$: $P(X=k) = \binom{4}{k}/16$. Sum $= (1+4+6+4+1)/16 = 16/16 = 1$ ✓.

5. $n = 20$, $p = 0.85$, $k = 17$. $P(X = 17) = \binom{20}{17}(0.85)^{17}(0.15)^3 = 1140 \cdot (0.85)^{17} \cdot (0.15)^3$.

Q1 (2 marks): $P(X = 4) = \binom{6}{4}(0.5)^4(0.5)^2 = 15 \cdot \tfrac{1}{16} \cdot \tfrac{1}{4}$ [1] $= \tfrac{15}{64}$ [1].

Q2 (3 marks): $X \sim \mathrm{Bin}(15, 0.25)$ [1]. $P(X = 5) = \binom{15}{5}(0.25)^5(0.75)^{10} = 3003 \cdot 0.0009766 \cdot 0.05631$ [1] $\approx 0.1651$ [1].

Q3 (3 marks): $X \sim \mathrm{Bin}(12, 0.7)$ [0.5]. $P(X \geq 10) = P(X=10) + P(X=11) + P(X=12)$ [0.5]. $P(X=10) = \binom{12}{10}(0.7)^{10}(0.3)^2 = 66 \cdot 0.02825 \cdot 0.09 \approx 0.1678$ [0.5]. $P(X=11) = 12 \cdot (0.7)^{11}(0.3) \approx 0.0712$ [0.5]. $P(X=12) = (0.7)^{12} \approx 0.0138$ [0.5]. Total $\approx 0.2528$ [0.5].

Boss battle · The Binomial Calculator

earn bronze · silver · gold

Five timed questions applying the binomial PMF: state $n$, $p$, $k$; compute $P(X=k)$; handle "at least" and "at most" via complements. 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 computing binomial probabilities. Lighter alternative to the boss.

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