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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 formulahow 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

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidencethat 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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