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Module 10 · L14 of 20 ~40 min ⚡ +90 XP available

Binomial Problems — Multi-Step

A pharmaceutical trial has 12 patients and a drug that works in 80% of cases. The funding body asks: what is the probability that at least 10 respond? That is no longer one term — it is a sum of $P(X=10) + P(X=11) + P(X=12)$, or equivalently $1 - P(X \leq 9)$. This lesson trains the strategies that turn binomial sums and complements into clean calculations.

Today's hook — A fair coin is flipped 4 times. Before reading on, write down two ways to express $P(\text{at least 1 head})$: (a) as a sum of individual probabilities, and (b) using the complement. Which is faster? Compare your answers after card 05.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

If $X \sim \text{Bin}(5, 0.5)$, write — without computing — two different ways to express $P(X \geq 4)$: one as a sum, one as a complement. Which would you choose for a calculator-free question?

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The two moves for multi-step binomial problems

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Every multi-step binomial problem rewards two habits: rewrite the event as a sum of $P(X=k)$ terms, then choose direct sum or complement — whichever has fewer terms. Picking the shorter route halves the work.

The sum-or-complement strategy: (1) list the values of $k$ in the event, (2) count them, (3) compare to $n$ + 1 minus that count — if the complement has fewer terms, use $1 - P(\text{complement})$.

$P(X \geq r) = \displaystyle\sum_{k=r}^{n}\binom{n}{k}p^k(1-p)^{n-k}$ · $P(X \geq 1) = 1 - P(X = 0)$

$P(X \geq 1) = 1 - P(X = 0)$

Translate the words first

"At least", "more than", "at most", "fewer than", "between" — each phrase maps to a set of $k$ values. Write the set explicitly before reaching for the formula.

Use the complement when it's shorter

"At least 1" out of $n=20$ trials is much faster as $1 - P(X=0)$ (one term) than summing $P(X=1) + \ldots + P(X=20)$ (twenty terms).

Watch the inequality boundaries

"More than 3" means $X \geq 4$, not $X \geq 3$. "Fewer than 5" means $X \leq 4$, not $X \leq 5$. Strict vs. non-strict inequalities are a common trap.

What you'll master

Know

Key facts

$P(X \leq r) = \displaystyle\sum_{k=0}^{r}\binom{n}{k}p^k(1-p)^{n-k}$ (cumulative)
$P(X \geq r) = 1 - P(X \leq r-1)$ (complement rule)
$P(a \leq X \leq b) = P(X \leq b) - P(X \leq a-1)$ (range)

Understand

Concepts

How to translate "at least", "at most", "more than", "fewer than" into ranges of $k$
Why the complement saves work when the event covers most of the distribution
How a multi-step problem chains identification, summing, and verification

Can do

Skills

Compute $P(X \leq r)$, $P(X \geq r)$, $P(a \leq X \leq b)$ for a binomial random variable
Choose between direct sum and complement to minimise the work
Apply the four-step strategy: identify, list, decide, compute

Key terms

Cumulative probability$P(X \leq r) = \displaystyle\sum_{k=0}^{r}\binom{n}{k}p^k(1-p)^{n-k}$ — the probability of getting up to and including $r$ successes.

Complement rule$P(\text{event}) = 1 - P(\text{not event})$. For binomial: $P(X \geq 1) = 1 - P(X = 0)$.

"At least $r$"$P(X \geq r)$. Includes $r$ itself and everything above. Often computed as $1 - P(X \leq r-1)$.

"At most $r$"$P(X \leq r)$. Includes $r$ itself and everything below. A direct cumulative sum.

"Between $a$ and $b$" (inclusive)$P(a \leq X \leq b) = P(X = a) + P(X = a+1) + \ldots + P(X = b)$, or $P(X \leq b) - P(X \leq a-1)$.

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

The four-step multi-step strategy

core concept

Multi-step binomial questions need a disciplined process:

Identify. Extract $n$, $p$ from the scenario as in lesson 13.
List. Write the set of $k$ values that make up the event. ("At least 3 out of 6" $\Rightarrow k \in \{3,4,5,6\}$.)
Decide. Count the terms. If the complement has fewer terms, use $1 - P(\text{complement})$.
Compute. Sum the $P(X=k)$ values from lesson 13's formula.

Worked through the hook: $X \sim \text{Bin}(4, 0.5)$, find $P(\text{at least 1 head})$:

Direct sum (4 terms): $P(X \geq 1) = P(X=1) + P(X=2) + P(X=3) + P(X=4)$.
Complement (1 term): $P(X \geq 1) = 1 - P(X=0) = 1 - \binom{4}{0}(0.5)^0(0.5)^4 = 1 - \tfrac{1}{16} = \tfrac{15}{16}$.
The complement wins easily — 1 term vs. 4 terms.

Phrase-to-range cheat sheet. "At least $r$": $X \geq r$. "More than $r$": $X \geq r+1$. "At most $r$": $X \leq r$. "Fewer than $r$": $X \leq r-1$. "Between $a$ and $b$ inclusive": $a \leq X \leq b$.

Multi-step binomial questions need a disciplined process:

Pause — copy the four-step multi-step strategy (identify, list $k$ values, decide complement, compute) with the hook example into your book.

Quick check: $X \sim \text{Bin}(10, 0.4)$. Which expression equals $P(X \geq 2)$?

Range probabilities and complement shortcuts

core concept

We just saw the four-step multi-step strategy: identify $n,p$; list the relevant $k$ values; decide whether the complement is shorter; compute. That raises a question: for interval events like $P(3\leq X\leq5)$, and for heavy-tailed events like $P(X\geq2)$, when is the complement always faster? This card answers it → use the complement when the excluded tail has fewer terms; $P(X\geq2)=1-P(X=0)-P(X=1)$ saves computing 6+ terms.

For ranges like $a \leq X \leq b$, sum the individual terms. For one-sided tails covering most of the distribution, the complement is faster.

Example 1 — range: $X \sim \text{Bin}(6, 0.5)$. Find $P(2 \leq X \leq 4)$.

$P(X=2) = \binom{6}{2}(0.5)^6 = 15/64$.
$P(X=3) = \binom{6}{3}(0.5)^6 = 20/64$.
$P(X=4) = \binom{6}{4}(0.5)^6 = 15/64$.
Sum $= 50/64 = 25/32 = 0.78125$.

Example 2 — complement: $X \sim \text{Bin}(20, 0.05)$. Find $P(X \geq 1)$.

Direct: 20 terms. Complement: 1 term.
$P(X \geq 1) = 1 - P(X=0) = 1 - (0.95)^{20} \approx 1 - 0.3585 = 0.6415$.

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

Common mistake. Forgetting to include the endpoint in "at most $r$": $P(X \leq r)$ includes $k = r$. Similarly "at least $r$" includes $k = r$. The boundary value is part of the event unless the wording says "strictly less than" or "more than".

For ranges like $a \leq X \leq b$, sum the individual terms. For one-sided tails covering most of the distribution, the complement is faster.

Pause — copy the complement shortcut rule: if the excluded tail has fewer terms than the event itself, use $1-P(\text{complement})$ — include the $P(X\geq2)$ example into your book.

Did you get this? True or false: for $X \sim \text{Bin}(n, p)$, $P(X > 3) = 1 - P(X \leq 3)$.

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

PROBLEM 1 · AT LEAST (COMPLEMENT)

A vaccine is effective in $90\%$ of patients. In a clinic of $15$ patients, find $P(\text{at least 14 respond})$.

Identify: $X \sim \text{Bin}(15, 0.9)$. Event: $X \geq 14$, so $k \in \{14, 15\}$. Only 2 terms — direct sum is the way.

Complement would need 14 terms ($k = 0, \ldots, 13$), so direct sum is clearly shorter here.

PROBLEM 2 · AT LEAST 1 (COMPLEMENT WIN)

A box contains light bulbs with a defect rate of $4\%$. A buyer inspects $25$ randomly selected bulbs. Find $P(\text{at least 1 is defective})$.

Identify: $X \sim \text{Bin}(25, 0.04)$. Event: $X \geq 1$, so $k \in \{1, 2, \ldots, 25\}$ (25 terms). Complement: $k = 0$ (1 term).

Massive difference — complement is one term vs. twenty-five. The complement always wins for "at least 1".

PROBLEM 3 · RANGE (DIRECT SUM)

A fair die is rolled $8$ times. Find $P(\text{between 2 and 4 sixes inclusive})$.

Identify: $X \sim \text{Bin}(8, 1/6)$. Event: $2 \leq X \leq 4$, so $k \in \{2, 3, 4\}$ (3 terms). Direct sum.

Range probabilities are most naturally computed as a direct sum, especially when only 2-4 terms are involved.

Fill the gap: If $X \sim \text{Bin}(4, 0.5)$, then $P(X \geq 1) = 1 - P(X = 0) = 1 - \dfrac{1}{} = \dfrac{15}{16}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Confusing "more than $r$" with "at least $r$"

"More than 3" means strictly greater: $X \geq 4$. "At least 3" includes 3: $X \geq 3$. These give different sums. Always re-read the question, underline the phrase, and write the inequality explicitly before listing $k$ values.

Trap 02

Using $1 - P(X \leq r)$ when the event is "at least $r$"

"At least $r$" is $X \geq r$, complement is $X \leq r - 1$. So $P(X \geq r) = 1 - P(X \leq r-1)$, NOT $1 - P(X \leq r)$. The latter is the complement of "at most $r$", a different event. The off-by-one error here is one of the most common HSC mistakes.

Trap 03

Forgetting to sum all required terms

For "between 2 and 5 inclusive", students sometimes compute only $P(X=2) + P(X=5)$, missing $k=3$ and $k=4$. Always list every $k$ value first, count them, then check your sum has the right number of terms.

Did you get this? True or false: for $X \sim \text{Bin}(10, 0.3)$, $P(X \geq 3) = 1 - P(X \leq 3)$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

$X \sim \text{Bin}(8, 0.5)$. Find $P(X \geq 6)$ as a direct sum.

A spinner gives "win" with probability $0.2$. In $30$ spins, use the complement to find $P(\text{at least 1 win})$.

A multiple-choice quiz has $10$ questions with $4$ options each. A student guesses every question. Find $P(\text{at most 2 correct})$.

A factory has a $6\%$ defect rate. A buyer samples $20$ items. Find $P(\text{more than 2 are defective})$.

$X \sim \text{Bin}(12, 0.4)$. Find $P(3 \leq X \leq 5)$ as a direct sum.

Odd one out: Three of these are correct ways to express $P(X \geq 2)$ for a binomial random variable $X \sim \text{Bin}(n, p)$. Which one is NOT?

Revisit your thinking

Earlier you wrote two expressions for $P(X \geq 4)$ when $X \sim \text{Bin}(5, 0.5)$.

As a sum: $P(X \geq 4) = P(X = 4) + P(X = 5) = \binom{5}{4}(0.5)^5 + \binom{5}{5}(0.5)^5 = (5 + 1)/32 = 6/32 = 3/16$. As a complement: $P(X \geq 4) = 1 - P(X \leq 3)$ — but $P(X \leq 3)$ has 4 terms ($k = 0, 1, 2, 3$), so the direct sum (2 terms) is faster. Picking the shorter path is the whole game.

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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 coin is flipped $6$ times. Find $P(\text{at least 5 heads})$. (2 marks)

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ApplyBand 43 marks

Q2. A factory has a $3\%$ defect rate. In a batch of $40$ items, find $P(\text{at least 1 is defective})$ using the complement. Round to 4 decimal places. (3 marks)

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AnalyseBand 53 marks

Q3. A drug is effective in $75\%$ of patients. In a trial of $10$ patients, find $P(\text{between 6 and 8 respond inclusive})$. Round to 4 decimal places. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $P(X \geq 6) = \binom{8}{6}(0.5)^8 + \binom{8}{7}(0.5)^8 + \binom{8}{8}(0.5)^8 = (28 + 8 + 1)/256 = 37/256 \approx 0.1445$.

2. $P(X \geq 1) = 1 - (0.8)^{30} \approx 1 - 0.001238 \approx 0.9988$.

3. $P(X \leq 2) = \binom{10}{0}(0.25)^0(0.75)^{10} + \binom{10}{1}(0.25)^1(0.75)^9 + \binom{10}{2}(0.25)^2(0.75)^8 \approx 0.0563 + 0.1877 + 0.2816 \approx 0.5256$.

4. $P(X > 2) = 1 - P(X \leq 2) = 1 - [P(X=0)+P(X=1)+P(X=2)]$ with $n=20, p=0.06$. $P(X=0) = (0.94)^{20} \approx 0.2901$; $P(X=1) = 20 \cdot 0.06 \cdot (0.94)^{19} \approx 0.3703$; $P(X=2) = \binom{20}{2}(0.06)^2(0.94)^{18} \approx 0.2246$. Sum $\approx 0.8850$. So $P(X > 2) \approx 0.1150$.

5. With $n=12, p=0.4$: $P(X=3) = \binom{12}{3}(0.4)^3(0.6)^9 \approx 0.1419$; $P(X=4) = \binom{12}{4}(0.4)^4(0.6)^8 \approx 0.2128$; $P(X=5) = \binom{12}{5}(0.4)^5(0.6)^7 \approx 0.2270$. Sum $\approx 0.5817$.

Q1 (2 marks): $P(X \geq 5) = \binom{6}{5}(0.5)^6 + (0.5)^6$ [1] $= (6+1)/64 = 7/64 \approx 0.1094$ [1].

Q2 (3 marks): $P(X \geq 1) = 1 - P(X=0)$ [1] $= 1 - (0.97)^{40}$ [1] $\approx 1 - 0.2957 \approx 0.7043$ [1].

Q3 (3 marks): $P(X=6) = \binom{10}{6}(0.75)^6(0.25)^4 \approx 0.1460$; $P(X=7) = \binom{10}{7}(0.75)^7(0.25)^3 \approx 0.2503$; $P(X=8) = \binom{10}{8}(0.75)^8(0.25)^2 \approx 0.2816$ [1 for any one term]; sum [1]; $\approx 0.6779$ [1].

Boss battle · The Binomial Multi-Step

earn bronze · silver · gold

Five timed questions on cumulative and complement binomial probabilities. 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 multi-step binomial questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 13 · Binomial Problems — Single Step Lesson 15 · Mean and Variance of Binomial →

Module overview · Extension 1 · Checkpoint 3