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Module 10 · L05 of 20 ~40 min ⚡ +90 XP available

Binomial Probability Formula

A pharmacist runs a trial where a new drug works in 70% of patients. If 8 patients are treated, what is the chance exactly 6 will respond? Brute-force listing every combination would take hours, but a single elegant formula handles it in one line. This lesson derives the binomial probability formula $P(X=k)=\binom{n}{k}p^k q^{n-k}$ and practises it with small $n$ before you scale up.

Today's hook, Flip a fair coin three times. Before reading on, list all eight outcomes and write down how many of them have exactly two heads. Then express that count as $\binom{3}{2}$ and check whether $\binom{3}{2}\cdot(\tfrac{1}{2})^2\cdot(\tfrac{1}{2})^1$ gives the probability you'd expect.

0/5QUESTS

Recall, your gut answer first

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A biased coin has $P(\text{heads}) = 0.6$. It is tossed 3 times. Before using any formulahow many distinct sequences contain exactly 2 heads, and what is the probability of any one such sequence? Write your reasoning below.

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The two moves for binomial problems

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Every binomial question rewards two habits: identify the trial (independent, two outcomes, fixed $n$ and $p$), then multiply count by probability using $\binom{n}{k}p^k q^{n-k}$ where $q = 1-p$. Naming $n$, $k$, $p$ and $q$ on the page is the most reliable way to avoid arithmetic slips.

The identify-then-substitute strategy: (1) confirm the trial is binomial (Bernoulli, fixed $n$, constant $p$, independent), (2) read off $n$, $k$, $p$ and $q$, (3) write the formula and substitute before evaluating.

$P(X = k) = \binom{n}{k} p^k q^{n-k}$ where $q = 1 - p$, $X \sim B(n, p)$.

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

Check the four BINS conditions

B inary outcomes, I ndependent trials, N fixed number of trials, S uccess probability $p$ constant. If any fails, the formula doesn't apply.

$q = 1 - p$ every time

Calculate $q$ explicitly. The exponent on $q$ is $n-k$, the number of failures. Mixing up $k$ and $n-k$ is the most common slip.

$\binom{n}{k}$ counts arrangements

The coefficient counts how many distinct ways $k$ successes can appear among $n$ trials. Without it you would only get the probability of a single ordered sequence.

What you'll master

Know

Key facts

Binomial probability: $P(X = k) = \displaystyle\binom{n}{k}p^k q^{n-k}$ with $q = 1 - p$
$X \sim B(n,p)$ means $X$ is binomial with parameters $n$ and $p$
The four BINS conditions for a binomial random variable

Understand

Concepts

Why the formula has three ingredients: a count, a success probability, and a failure probability
How $\binom{n}{k}$ comes from counting the arrangements of $k$ successes in $n$ trials
Why the trials must be independent and $p$ must stay constant

Can do

Skills

Substitute small values of $n$, $k$, $p$ into the formula correctly
Recognise a binomial scenario from a worded problem
Derive the formula by combining the multiplication and addition principles

Key terms

Bernoulli trialA single experiment with exactly two outcomes, usually labelled "success" (probability $p$) and "failure" (probability $q = 1 - p$).

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

$\binom{n}{k}$The binomial coefficient $\dfrac{n!}{k!(n-k)!}$, equal to the number of ways to choose $k$ successes from $n$ positions.

Success probability $p$The probability of the outcome being counted by $X$. Must remain constant across all $n$ trials.

Failure probability $q$$q = 1 - p$, the probability of the complementary outcome. Appears with exponent $n - k$ in the formula.

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

Where the formula comes from

core concept

The binomial formula is built from two basic principles you already know, the multiplication principle (independent events multiply) and the counting principle (count the arrangements).

One ordered sequence. A single sequence with $k$ successes and $n-k$ failures (in any specific order) has probability $p^k q^{n-k}$ because the trials are independent.
Count the arrangements. The number of ways to choose which $k$ of the $n$ trials are the successes is $\binom{n}{k}$.
Combine. Each arrangement is mutually exclusive, so we add their probabilities, giving $\binom{n}{k}p^k q^{n-k}$.

Worked through the hook: Three tosses of a fair coin, $p = q = \tfrac{1}{2}$, $n = 3$, $k = 2$.

Sequences with exactly 2 heads: HHT, HTH, THH, that's $\binom{3}{2} = 3$ arrangements.
Each sequence: probability $(\tfrac{1}{2})^2 \cdot (\tfrac{1}{2})^1 = \tfrac{1}{8}$.
Total: $P(X = 2) = 3 \cdot \tfrac{1}{8} = \tfrac{3}{8}$.
Check by formula: $\binom{3}{2}(\tfrac{1}{2})^2(\tfrac{1}{2})^1 = 3 \cdot \tfrac{1}{4} \cdot \tfrac{1}{2} = \tfrac{3}{8}$ ✓.

Why the count matters. Without the $\binom{n}{k}$ factor, you would only be counting one specific sequence, say HHT, and ignoring HTH and THH. The coefficient bundles all valid arrangements into a single number.

The binomial formula is built from two basic principles you already know, the multiplication principle (independent events multiply) and the counting principle (count the arrangements).

Pause, copy the derivation of the binomial formula: (1) probability of one ordered sequence $= p^k(1-p)^{n-k}$; (2) multiply by $\binom{n}{k}$ arrangements into your book.

Quick check: A biased die shows a six with probability $\tfrac{1}{4}$. It is rolled 4 times. Which expression gives the probability of exactly 2 sixes?

Practising with small $n$

core concept

We just saw that the binomial formula combines the multiplication principle ($p^k(1-p)^{n-k}$ for one ordered sequence) and the counting principle ($\binom{n}{k}$ arrangements). That raises a question: for small $n$ (say $n=4$), can you verify the formula by listing all outcomes explicitly, and is the result the same? This card answers it → yes: for $n=4$, $k=1$, $p=0.3$, listing the four FSSS-type sequences gives $4\times0.3\times0.7^3=\binom{4}{1}(0.3)^1(0.7)^3$.

Small values of $n$ (say $n \leq 5$) let you double-check the formula by listing outcomes. This sanity check builds confidence before you move on to larger $n$ where listing is impractical.

Example: $n = 4$ tosses of a biased coin with $p = 0.3$. Find $P(X = 1)$.

$n = 4$, $k = 1$, $p = 0.3$, $q = 0.7$.
$\binom{4}{1} = 4$ (the single head can occupy any of 4 positions).
$P(X = 1) = 4 \cdot (0.3)^1 \cdot (0.7)^3 = 4 \cdot 0.3 \cdot 0.343 = 0.4116$.

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

Sanity check. Adding $P(X = 0) + P(X = 1) + \dots + P(X = n)$ must give 1. For $n = 4$, $p = 0.3$: $(0.7)^4 + 4(0.3)(0.7)^3 + 6(0.3)^2(0.7)^2 + 4(0.3)^3(0.7) + (0.3)^4 = 1$ ✓. This is just the binomial expansion of $(p + q)^n = 1^n = 1$.

Small values of $n$ (say $n \leq 5$) let you double-check the formula by listing outcomes. This sanity check builds confidence before you move on to larger $n$ where listing is impractical.

Pause, copy the small-$n$ verification method: list all sequences explicitly for $n=4$, $k=1$ and confirm the count matches $\binom{4}{1}=4$ into your book.

Did you get this? True or false: if $X \sim B(5, 0.4)$, then $P(X = 3) = \binom{5}{3}(0.4)^3 (0.6)^2$.

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

PROBLEM 1 · DIRECT SUBSTITUTION

A fair die is rolled 5 times. Find the probability of rolling exactly 2 sixes.

Identify: $n = 5$, $k = 2$, $p = \tfrac{1}{6}$ (rolling a six), $q = 1 - \tfrac{1}{6} = \tfrac{5}{6}$.

Confirm BINS: 5 trials (N fixed), each roll independent (I), six or not-six (B), $p = \tfrac{1}{6}$ each time (S). Binomial applies.

PROBLEM 2 · WORDED CONTEXT

In a multiple-choice quiz, each question has 4 options with exactly one correct answer. A student guesses all 6 questions. Find the probability they answer exactly 3 correctly.

Identify: $n = 6$, $k = 3$, $p = \tfrac{1}{4}$ (guess correctly), $q = \tfrac{3}{4}$.

Each question is an independent Bernoulli trial. The student's guesses don't influence one another, so trials are independent and $p$ stays constant.

PROBLEM 3 · ALL-SUCCESS / ALL-FAILURE

A test for a disease is 92% accurate (gives the correct result 92% of the time). It is independently applied to 4 patients. Find the probability that all 4 results are correct.

Identify: $n = 4$, $k = 4$, $p = 0.92$, $q = 0.08$. We want $P(X = 4)$.

"All correct" means $k = n$, so the formula simplifies because $\binom{n}{n} = 1$ and $q^{n-k} = q^0 = 1$.

Fill the gap: If $X \sim B(7, 0.4)$, then the probability of exactly 2 successes is $P(X = 2) = \binom{7}{}(0.4)^2(0.6)^{}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the $\binom{n}{k}$ factor

Writing just $p^k q^{n-k}$ gives the probability of one specific ordered sequence, not the probability of "exactly $k$ successes". Always include the $\binom{n}{k}$ coefficient to count all arrangements with $k$ successes.

Trap 02

Swapping the exponents on $p$ and $q$

$p$ is raised to $k$ (the number of successes) and $q$ to $n - k$ (the number of failures). Students often write $p^{n-k} q^k$ by mistake. Label each exponent on the page before computing.

Trap 03

Applying the formula when the trials aren't independent

Drawing cards without replacement, or scenarios where $p$ changes between trials, break the binomial conditions. If trials aren't independent or $p$ shifts, you must use conditional probability or hypergeometric methods instead.

Did you get this? True or false: drawing 5 cards without replacement from a standard deck and counting the number of hearts is a binomial random variable.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A fair coin is tossed 5 times. Find the probability of exactly 3 heads.

If $X \sim B(6, 0.2)$, find $P(X = 4)$. Give your answer to 4 decimal places.

A biased coin has $P(\text{head}) = \tfrac{2}{3}$. Find the probability of exactly 4 heads in 5 tosses.

A factory makes light bulbs with a 5% defect rate. In a batch of 4 bulbs, find the probability that none are defective.

Derive (don't just quote) the probability that, in $n = 3$ independent trials of a Bernoulli experiment with success probability $p$, exactly 2 successes occur. Show how the count $\binom{3}{2}$ and the single-sequence probability combine.

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

Revisit your thinking

Earlier you considered three tosses of a biased coin with $p = 0.6$ and counted the sequences with exactly 2 heads.

The three sequences HHT, HTH, THH each have probability $(0.6)^2(0.4) = 0.144$, so $P(X = 2) = 3 \cdot 0.144 = 0.432$. The "3" is exactly $\binom{3}{2}$, the binomial coefficient is doing the counting for you. Once you see that pattern, the formula stops feeling abstract.

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Multiple choice

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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. A fair coin is tossed 4 times. Find the probability of exactly 3 heads. (2 marks)

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ApplyBand 43 marks

Q2. A test is 80% accurate. It is independently applied to 5 patients. Find the probability that exactly 4 results are correct, leaving your answer in exact form. (3 marks)

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AnalyseBand 53 marks

Q3. A biased die shows a one with probability $\tfrac{1}{5}$. The die is rolled 6 times. Derive (do not just quote) an expression for the probability of exactly 2 ones using the multiplication and addition principles, then evaluate. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $n=5, k=3, p=q=\tfrac{1}{2}$. $P = \binom{5}{3}(\tfrac{1}{2})^3(\tfrac{1}{2})^2 = 10 \cdot \tfrac{1}{32} = \tfrac{5}{16}$.

2. $P(X=4) = \binom{6}{4}(0.2)^4(0.8)^2 = 15 \cdot 0.0016 \cdot 0.64 = 0.01536 \approx 0.0154$.

3. $P = \binom{5}{4}(\tfrac{2}{3})^4(\tfrac{1}{3})^1 = 5 \cdot \tfrac{16}{81} \cdot \tfrac{1}{3} = \tfrac{80}{243} \approx 0.329$.

4. $P(X=0) = (0.95)^4 = 0.81450625 \approx 0.8145$.

5. The three sequences are SSF, SFS, FSS, each with probability $p^2 q$ (multiplication, independent). They are mutually exclusive, so by addition $P = 3 p^2 q = \binom{3}{2} p^2 q^1$. This is the binomial formula with $n=3, k=2$.

Q1 (2 marks): $P(X=3) = \binom{4}{3}(\tfrac{1}{2})^3(\tfrac{1}{2})^1$ [1] $= 4 \cdot \tfrac{1}{16} = \tfrac{1}{4}$ [1].

Q2 (3 marks): $n=5, k=4, p=0.8, q=0.2$ [1]. $P(X=4) = \binom{5}{4}(0.8)^4(0.2)^1$ [1] $= 5 \cdot 0.4096 \cdot 0.2 = 0.4096 = \tfrac{256}{625}$ [1].

Q3 (3 marks): One specific ordered sequence with 2 ones and 4 non-ones has probability $(\tfrac{1}{5})^2(\tfrac{4}{5})^4$ by independence [1]. The number of such arrangements is $\binom{6}{2} = 15$ [1]. By the addition principle: $P = 15 \cdot \tfrac{1}{25} \cdot \tfrac{256}{625} = \tfrac{3840}{15625} = \tfrac{768}{3125} \approx 0.246$ [1].

Boss battle · The Binomial Builder

earn bronze · silver · gold

Five timed questions applying the binomial probability formula to fresh scenarios. 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 binomial probability questions. Lighter alternative to the boss.

Mark lesson as complete

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Module overview · Extension 1 · Checkpoint 1