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Module 10 · L06 of 20 ~40 min ⚡ +90 XP available

Calculating Binomial Probabilities

You know the formula $P(X=k) = \binom{n}{k}p^k q^{n-k}$. Now we make the arithmetic painless: an ordered keystroke routine, when to keep exact fractions vs. when to round, and how to spot rounding errors before they cost marks. By the end you can evaluate a binomial probability in under 60 seconds with confidence in the answer.

Today's hook — Without picking up a calculator, predict whether $P(X = 5)$ for $X \sim B(8, 0.7)$ is closer to $0.05$, $0.25$ or $0.50$. Note your guess. After this lesson you'll know how to compute the exact answer in three keystrokes and check whether your intuition was on the mark.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

$X \sim B(8, 0.7)$. Before reaching for a calculator — write out the formula for $P(X = 5)$ with $n$, $k$, $p$ and $q$ labelled, then predict whether the answer is roughly 0.05, 0.25 or 0.50. Justify your guess.

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The two moves for efficient evaluation

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Efficient binomial calculation rewards two habits: write the substituted formula in full first, then use the nCr key rather than computing factorials by hand. Going straight to the calculator without writing the formula is the leading cause of "lost" method marks.

The write-then-press routine: (1) substitute $n, k, p, q$ into $\binom{n}{k}p^k q^{n-k}$ on paper, (2) read off the three factors and enter them one at a time, (3) decide exact or decimal based on the question wording.

Keystrokes: n nCr k × p ^ k × q ^ (n−k) =

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

Use the nCr key

Calculators have a dedicated nCr button (often SHIFT + a function key). It's faster and more accurate than computing $\dfrac{n!}{k!(n-k)!}$ by hand.

Exact or decimal?

If the question says "leave your answer in exact form" or uses simple fractions, keep fractions. If it asks for "to 4 decimal places" or uses decimal $p$, convert.

Estimate before evaluating

Mode (most likely value) of $B(n, p)$ is near $np$. If $np = 5.6$, then $P(X = 5)$ and $P(X = 6)$ should be the largest probabilities — a useful sanity check.

What you'll master

Know

Key facts

Calculator keystroke order: nCr · ^ · ^ · =
Exact form keeps fractions; decimal form rounds to stated precision
Mode of $B(n, p) \approx np$ — used as a sanity check

Understand

Concepts

Why writing the substituted formula first protects method marks
When exact form is required vs. when a decimal is fine
How rounding too early (especially squaring rounded values) propagates error

Can do

Skills

Evaluate $\binom{n}{k}p^k q^{n-k}$ on a calculator in under 60 seconds
Convert between exact and decimal answers cleanly
Estimate the rough size of $P(X = k)$ before computing

Key terms

Exact formAn answer expressed as a fraction or unevaluated power (e.g., $\tfrac{135}{1024}$ or $20 \cdot (0.25)^3(0.75)^3$) rather than a decimal approximation.

Decimal formAn answer rounded to a stated number of decimal places or significant figures, used when the question asks for it or $p$ is itself a decimal.

nCr keyThe calculator function that returns $\binom{n}{k}$ directly, usually accessed via SHIFT on Casio and similar models.

Mode of $B(n,p)$The most probable value of $X$. Approximately $\lfloor (n+1)p \rfloor$; in practice $\approx np$. Useful for predicting which $P(X = k)$ will be largest.

Premature roundingRounding $p^k$ or $q^{n-k}$ before multiplying — magnifies error. Always keep full precision until the final step.

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

The keystroke routine

core concept

A reliable workflow with any scientific calculator:

Write the formula with values substituted. Example: $P(X = 5) = \binom{8}{5}(0.7)^5(0.3)^3$.
Enter $\binom{n}{k}$. Type $n$, press nCr (or SHIFT + a function key), type $k$. Press ×.
Enter $p^k$. Type $p$, press ^, type $k$. Press ×.
Enter $q^{n-k}$. Type $q$, press ^, type $n - k$. Press =.
Convert if required. If exact form is needed, use the fraction/decimal toggle (often S⇔D).

Worked through the hook: $X \sim B(8, 0.7)$, $P(X = 5)$.

Substituted formula: $P(X=5) = \binom{8}{5}(0.7)^5(0.3)^3$.
$\binom{8}{5} = 56$, $(0.7)^5 = 0.16807$, $(0.3)^3 = 0.027$.
Product: $56 \cdot 0.16807 \cdot 0.027 = 0.254\ldots$ — so closer to $0.25$ than $0.05$ or $0.50$.

Why the order matters. Calculating $\binom{n}{k}$ first gives a tidy integer (here, 56) that anchors the rest of the arithmetic. Computing $p^k q^{n-k}$ first produces a tiny decimal that's harder to verify visually.

Calculator routine: (1) Write $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$ with values. (2) Enter nCr. (3) Multiply by $p^k$. (4) Multiply by $(1-p)^{n-k}$. (5) Convert to required form.

Pause — copy the five-step calculator keystroke routine for evaluating $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$ into your book.

Quick check: $X \sim B(10, 0.5)$. To find $P(X = 4)$ efficiently, the best calculator entry is:

Exact form vs. decimal answers

core concept

We just saw the calculator keystroke routine: write the formula, enter $\binom{n}{k}$ via nCr, multiply by $p^k$ and $(1-p)^{n-k}$. That raises a question: when should you leave the answer as an exact fraction and when should you convert to a decimal — and how does the wording of the question signal which is expected? This card answers it → "exact form" or a "nice" $p$ like $\frac{1}{2}$ signals a fraction; a decimal $p$ like $0.73$ signals a 4-decimal-place answer.

HSC binomial questions usually fall into one of two buckets, signalled by the wording and the value of $p$:

Exact form expected — when $p$ is a "nice" fraction like $\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \tfrac{1}{6}$ or when the question says "exact form" or "as a fraction". Keep everything as fractions.
Decimal expected — when $p$ is given as a decimal (e.g. $p = 0.7$, $p = 0.92$) or the question says "to 4 decimal places" / "to 3 significant figures".

Example (exact): $X \sim B(5, \tfrac{1}{3})$, $P(X = 2) = \binom{5}{2}(\tfrac{1}{3})^2(\tfrac{2}{3})^3 = 10 \cdot \tfrac{1}{9} \cdot \tfrac{8}{27} = \tfrac{80}{243}$.

Example (decimal): $X \sim B(10, 0.4)$, $P(X = 3) = \binom{10}{3}(0.4)^3(0.6)^7 = 120 \cdot 0.064 \cdot 0.0279936 = 0.21499 \approx 0.2150$ (4 d.p.).

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

Premature rounding warning. If you round $(0.7)^5$ to $0.17$ before multiplying, the final answer will be off in the 3rd decimal — enough to lose a mark in a 4 d.p. answer. Keep full precision on the calculator until the very last step, then round.

HSC binomial questions usually fall into one of two buckets, signalled by the wording and the value of $p$:

Pause — copy the decision rule: use exact fractions when $p$ is a simple fraction or the question says "exact form"; use 4 d.p. decimals when $p$ is given as a decimal into your book.

Did you get this? True or false: if the question asks for $P(X=2)$ where $X \sim B(4, \tfrac{1}{2})$ and the wording specifies "exact form", the correct answer is $\tfrac{3}{8}$.

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

PROBLEM 1 · DECIMAL ANSWER, 4 d.p.

$X \sim B(8, 0.7)$. Find $P(X = 5)$, giving your answer correct to 4 decimal places.

Identify: $n = 8$, $k = 5$, $p = 0.7$, $q = 0.3$. Write: $P(X = 5) = \displaystyle\binom{8}{5}(0.7)^5(0.3)^3$.

Always write the substituted formula on paper before pressing any keys — this secures the method mark even if the arithmetic slips.

PROBLEM 2 · EXACT FORM

$X \sim B(6, \tfrac{1}{3})$. Find $P(X = 2)$ in exact form.

Identify: $n = 6$, $k = 2$, $p = \tfrac{1}{3}$, $q = \tfrac{2}{3}$. Write: $P(X = 2) = \displaystyle\binom{6}{2}\left(\tfrac{1}{3}\right)^2\left(\tfrac{2}{3}\right)^4$.

Because $p$ is a simple fraction, keep everything as fractions. Convert to a decimal only if the question asks.

PROBLEM 3 · ESTIMATE THEN COMPUTE

$X \sim B(20, 0.15)$. Estimate which value of $k$ gives the largest $P(X = k)$, then compute $P(X = 3)$ to 4 decimal places.

Mode estimate: $np = 20 \cdot 0.15 = 3$. So $P(X = 3)$ should be the largest probability.

Knowing the mode in advance protects against silly mistakes — if the calculator returns 0.001 for $P(X = 3)$, that contradicts the estimate and signals an entry error.

Fill the gap: $X \sim B(6, \tfrac{1}{3})$. Then $P(X = 2) = \binom{6}{2}\left(\tfrac{1}{3}\right)^2\left(\tfrac{2}{3}\right)^4 = \dfrac{}{243}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Premature rounding

Rounding $(0.7)^5$ to $0.17$ before multiplying gives $56 \cdot 0.17 \cdot 0.027 = 0.2570$, which rounds incorrectly to $0.2570$ instead of $0.2541$. Keep every intermediate value at full calculator precision and round only on the final line.

Trap 02

Returning a decimal when exact form is required

"Express in exact form" rules out decimal approximations. If the question gives $p$ as a fraction (e.g. $\tfrac{1}{4}$) and asks for exact form, leave your answer as a fraction such as $\tfrac{135}{1024}$. Writing $0.132$ loses an answer mark.

Trap 03

Mis-entering parentheses on the calculator

Typing 56 × 0.7 ^ 5 × 0.3 ^ 3 without parentheses around the powers can give wrong results on some calculators because of operator precedence. When in doubt, wrap each power: 56 × (0.7 ^ 5) × (0.3 ^ 3) =.

Did you get this? True or false: for $X \sim B(12, 0.2)$, the mode (most probable value of $X$) is approximately 2.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

$X \sim B(10, 0.5)$. Find $P(X = 4)$ correct to 4 decimal places.

$X \sim B(5, \tfrac{1}{4})$. Find $P(X = 3)$ in exact form.

$X \sim B(15, 0.3)$. Find $P(X = 5)$ correct to 4 decimal places. Compare with $np$ to check whether 5 is near the mode.

$X \sim B(7, \tfrac{2}{5})$. Find $P(X = 3)$ first in exact form, then as a decimal to 4 d.p. Confirm the two agree.

Show that for $X \sim B(8, 0.5)$, $P(X = 3) = P(X = 5)$. Justify using the formula (without computing each value).

Odd one out: Three of these are valid ways to evaluate $P(X = 3)$ for $X \sim B(5, 0.4)$. Which is NOT?

Revisit your thinking

Earlier you guessed whether $P(X = 5)$ for $X \sim B(8, 0.7)$ is closer to $0.05$, $0.25$ or $0.50$.

The exact value is $\binom{8}{5}(0.7)^5(0.3)^3 = 56 \cdot 0.16807 \cdot 0.027 \approx 0.2541$ — closest to $0.25$. The mode is near $np = 5.6$, so $P(X = 5)$ and $P(X = 6)$ should be the two largest probabilities. Even with no formula in hand, $np$ gives a fast estimate of where the distribution is concentrated.

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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. $X \sim B(6, \tfrac{1}{2})$. Find $P(X = 4)$ in exact form. (2 marks)

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ApplyBand 43 marks

Q2. $X \sim B(12, 0.25)$. Find $P(X = 4)$ correct to 4 decimal places. Show the substituted formula. (3 marks)

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AnalyseBand 53 marks

Q3. $X \sim B(10, p)$ and $P(X = 4) = P(X = 6)$. Without computing decimals, deduce the value of $p$ and justify your reasoning. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $P(X=4) = \binom{10}{4}(0.5)^{10} = 210 \cdot \tfrac{1}{1024} = \tfrac{210}{1024} = \tfrac{105}{512} \approx 0.2051$.

2. $P(X=3) = \binom{5}{3}(\tfrac{1}{4})^3(\tfrac{3}{4})^2 = 10 \cdot \tfrac{1}{64} \cdot \tfrac{9}{16} = \tfrac{90}{1024} = \tfrac{45}{512}$.

3. $np = 4.5$, so mode is 4 or 5. $P(X=5) = \binom{15}{5}(0.3)^5(0.7)^{10} = 3003 \cdot 0.00243 \cdot 0.02825 \approx 0.2061$. Yes, $k=5$ is near the mode.

4. Exact: $\binom{7}{3}(\tfrac{2}{5})^3(\tfrac{3}{5})^4 = 35 \cdot \tfrac{8}{125} \cdot \tfrac{81}{625} = \tfrac{22680}{78125} = \tfrac{4536}{15625}$. Decimal: $\approx 0.2903$.

5. $P(X=3) = \binom{8}{3}(0.5)^3(0.5)^5 = 56(0.5)^8$. $P(X=5) = \binom{8}{5}(0.5)^5(0.5)^3 = 56(0.5)^8$. Since $\binom{8}{3} = \binom{8}{5} = 56$ and the product of powers $= (0.5)^8$ in both cases, the probabilities are equal.

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

Q2 (3 marks): $n=12, k=4, p=0.25, q=0.75$. $P(X=4) = \binom{12}{4}(0.25)^4(0.75)^8$ [1] $= 495 \cdot 0.00390625 \cdot 0.10011\ldots$ [1] $\approx 0.1936$ (4 d.p.) [1].

Q3 (3 marks): $\binom{10}{4} = \binom{10}{6} = 210$ [1]. So $P(X=4) = P(X=6)$ becomes $210 p^4 q^6 = 210 p^6 q^4 \Rightarrow p^4 q^4 (q^2 - p^2) = 0$ [1]. Since $p, q \in (0,1)$, $p^4q^4 \neq 0$, so $p^2 = q^2$. Taking positive roots: $p = q = \tfrac{1}{2}$ [1].

Boss battle · The Binomial Calculator

earn bronze · silver · gold

Five timed binomial probability evaluations — a mix of exact and decimal answers. 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 evaluating binomial probabilities under time pressure. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 05 · Binomial Probability Formula Lesson 07 · Cumulative Binomial Probabilities →

Module overview · Extension 1 · Checkpoint 2