Your weak spots

Insights load after your first practice round.

Module 10 · L07 of 20 ~40 min ⚡ +90 XP available

Cumulative Binomial Probabilities

A factory inspects batches of 20 widgets where each widget has a 6% chance of being defective. The shipment is rejected if there are more than 3 defectives in the sample. What is the probability of rejection? Questions like this rarely ask for a single $P(X = k)$ — they ask for cumulative ranges such as $P(X \leq 3)$ or $P(X \geq 4)$. This lesson trains you to sum binomial probabilities efficiently and use the complement trick to keep the arithmetic manageable.

Today's hook — A fair coin is tossed 5 times. Before reading on, write down which is easier to compute: $P(X \geq 1)$ (at least one head) by direct sum, or by using $1 - P(X = 0)$? Then estimate the probability mentally. Compare your reasoning after card 05.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

If $X \sim B(n,p)$, the probability of exactly $k$ successes is $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$. Before any formula — write down how you would compute $P(X \leq 2)$ for $X \sim B(5, 0.3)$. Which individual probabilities would you sum, and in which order would you compute them?

auto-saved

The two moves for cumulative probability

+5 XP to read

Every cumulative binomial calculation rewards two habits: decode the inequality precisely (does "at least", "at most" or "more than" mean $\leq$, $\geq$, $>$ or $<$?), then choose direct sum or complement to minimise the number of terms.

The decode-then-shortcut strategy: (1) translate the words to an inequality on $X$, (2) count how many terms each approach needs, (3) use the complement if it cuts the work in half.

Direct: $P(X \leq k)=\sum_{i=0}^{k}\binom{n}{i}p^i(1-p)^{n-i}$ · Complement: $P(X \geq k)=1-P(X \leq k-1)$

$P(X \geq k) = 1 - P(X \leq k-1)$

Watch strict vs weak

"$X > 3$" means $X \geq 4$ (since $X$ is integer-valued); "at most 3" is $X \leq 3$; "more than 3" excludes $3$ itself. A single misread shifts every term.

Sum what's shortest

If the question asks $P(X \geq 4)$ for $n = 20$, summing 17 terms is silly. Use $1 - P(X \leq 3)$ — only 4 terms.

Probabilities between

$P(a \leq X \leq b) = P(X \leq b) - P(X \leq a-1)$, or just sum $P(X=k)$ for $k = a,\dots,b$ when the range is small.

What you'll master

Know

Key facts

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

Understand

Concepts

Why integer-valued $X$ makes "$>$" equivalent to "$\geq$ next integer"
When the complement trick saves time (right-tail probabilities)
Why a difference of two cumulatives gives an interval probability

Can do

Skills

Translate word problems into the correct inequality on $X$
Compute $P(X \leq k)$, $P(X \geq k)$, $P(a \leq X \leq b)$ by direct sum
Use the complement to minimise arithmetic in right-tail questions

Key terms

Cumulative probability$P(X \leq k)$, the sum of $P(X = i)$ for all $i$ from $0$ up to $k$. Always lies between $0$ and $1$ inclusive.

Complement$P(\overline{A}) = 1 - P(A)$. For the binomial: $P(X \geq k) = 1 - P(X \leq k-1)$.

Right tailProbability $P(X \geq k)$ for $k$ close to $n$. Almost always computed via the complement.

Left tailProbability $P(X \leq k)$ for $k$ close to $0$. Direct summation is usually fastest here.

Interval probability$P(a \leq X \leq b)$. Compute either as a small sum or as $P(X \leq b) - P(X \leq a-1)$.

ME12-5NESA outcome: applies appropriate statistical processes including the binomial distribution and the normal distribution to analyse and solve problems.

The cumulative strategy

core concept

Cumulative binomial questions almost never ask for a single $P(X=k)$. The four-step approach always works:

Decode. Translate the words ("at least", "fewer than", "more than", "no more than") into an inequality.
Count. How many terms does direct summation need? How many for the complement?
Sum. Use whichever has fewer terms. Write the formula, then evaluate.
Verify. Check the answer lies in $[0,1]$ and matches intuition (small $p$ → small mass on large $k$).

Worked through the hook: Fair coin, $X \sim B(5, 0.5)$, find $P(X \geq 1)$:

Direct sum: $P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)$ — 5 terms.
Complement: $1 - P(X=0) = 1 - (0.5)^5 = 1 - \dfrac{1}{32} = \dfrac{31}{32}$ — 1 term.
The complement is dramatically faster for right-tail questions.

The integer-valued trick. Because $X$ takes only integer values, "$X > 3$" is the same as "$X \geq 4$". Always rewrite strict inequalities as weak ones on the next integer before applying the cumulative formula.

Cumulative strategy: decode inequality ("at least $k$" $\Rightarrow P(X\geq k)=1-P(X\leq k-1)$), count terms, use complement if shorter, verify answer in $[0,1]$.

Pause — copy the four-step cumulative strategy (decode, count, sum, verify) and the four inequality translations ("at least", "at most", "more than", "fewer than") into your book.

Quick check: Let $X \sim B(10, 0.2)$. Which expression correctly gives $P(X \geq 3)$ using the complement trick?

Interval probabilities and decoding language

core concept

We just saw the four-step cumulative strategy: decode language into an inequality, count how many terms each approach needs, sum the fewer ones, then verify the result lies in $[0,1]$. That raises a question: how exactly do you decode "between 3 and 5 inclusive" into a sum, and how many terms does the complement save for $B(8,0.5)$? This card answers it → $P(3\leq X\leq5)=P(X=3)+P(X=4)+P(X=5)$; the complement has 6 terms, so direct summation is faster here.

Words matter. Translate carefully:

"at most $k$" → $P(X \leq k)$
"at least $k$" → $P(X \geq k) = 1 - P(X \leq k-1)$
"more than $k$" → $P(X > k) = P(X \geq k+1) = 1 - P(X \leq k)$
"fewer than $k$" → $P(X < k) = P(X \leq k-1)$
"between $a$ and $b$ inclusive" → $P(a \leq X \leq b)$

Example: $X \sim B(8, 0.5)$. Find $P(3 \leq X \leq 5)$.

$P(X=3) = \binom{8}{3}(0.5)^8 = 56/256$
$P(X=4) = \binom{8}{4}(0.5)^8 = 70/256$
$P(X=5) = \binom{8}{5}(0.5)^8 = 56/256$
Sum: $182/256 = 91/128 \approx 0.711$.

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

Common mistake. Don't confuse $P(X < k)$ with $P(X \leq k)$ — they differ by exactly $P(X = k)$. Read the inequality symbol carefully before writing any summation.

Example: $X \sim B(8, 0.5)$. Find $P(3 \leq X \leq 5)$.

Pause — copy the interval probability formula $P(a\leq X\leq b)=\sum_{k=a}^b P(X=k)$ and the complement shortcut $P(X\geq k)=1-P(X\leq k-1)$ into your book.

Did you get this? True or false: for an integer-valued random variable $X$, "$P(X > 5)$" and "$P(X \geq 6)$" describe the same probability.

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

PROBLEM 1 · "AT LEAST ONE" SHORTCUT

A biased coin lands heads with probability $0.3$. It is tossed $6$ times. Find the probability of at least one head. Give the answer to 4 decimal places.

Let $X$ = number of heads. Then $X \sim B(6, 0.3)$ and we want $P(X \geq 1)$.

"At least one" decodes to $X \geq 1$. Counting terms: direct sum needs $6$ terms ($k=1,\dots,6$), complement needs only $1$ ($k=0$).

PROBLEM 2 · DIRECT CUMULATIVE SUM

A multiple-choice quiz has $10$ questions, each with $4$ options. A student guesses every answer. Find the probability of getting at most 2 correct.

Let $X$ = number correct. Each guess succeeds with probability $0.25$, so $X \sim B(10, 0.25)$. Want $P(X \leq 2)$.

"At most 2" is the left tail, so direct summation of $3$ terms ($k=0,1,2$) is the natural choice.

PROBLEM 3 · INTERVAL VIA DIFFERENCE OF CUMULATIVES

A factory produces components, $5\%$ of which are defective. A batch of $20$ is inspected. Find the probability that the number of defectives is between $2$ and $4$ inclusive.

Let $X$ = number of defectives. $X \sim B(20, 0.05)$. Want $P(2 \leq X \leq 4)$.

Three terms either way ($k=2,3,4$); direct sum is fine here. Could also use $P(X \leq 4) - P(X \leq 1)$.

Fill the gap: For $X \sim B(5, 0.5)$, the probability of getting at least one head is $P(X \geq 1) = 1 - (0.5)^5 = \dfrac{}{32}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Confusing strict and weak inequalities

"More than 3" means $X \geq 4$, not $X \geq 3$. Including the boundary value adds an extra term and gives a wrong answer that is off by exactly $P(X = 3)$. Always rewrite strict to weak on the next integer.

Trap 02

Wrong complement: $1 - P(X \leq k)$ vs $1 - P(X \leq k-1)$

$P(X \geq k) = 1 - P(X \leq k-1)$, NOT $1 - P(X \leq k)$. The second one gives $P(X \geq k+1)$. The difference is exactly $P(X = k)$ — easy to miss if you don't check whether the boundary is included.

Trap 03

Computing $(0.95)^{18}$ approximations carelessly

Cumulative sums involve products of small probabilities and large binomial coefficients. Truncating $(0.95)^{18}$ too aggressively can throw the final sum off by 5% or more. Carry at least 4 decimal places in intermediate steps and round only at the end.

Did you get this? True or false: $P(X \geq 4) = 1 - P(X \leq 4)$ for any binomial $X$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

$X \sim B(8, 0.4)$. Find $P(X \leq 2)$ by direct summation. Show all three terms.

A fair die is rolled $12$ times. Find the probability of getting at least one six. Use the complement trick.

In a shipment of $15$ widgets, $10\%$ are defective. Find the probability that more than 2 are defective. Choose direct sum or complement and justify.

$X \sim B(10, 0.5)$. Find $P(4 \leq X \leq 6)$. Compute as both a direct sum and as $P(X \leq 6) - P(X \leq 3)$, and check they agree.

A basketball player makes $70\%$ of free throws. In $10$ attempts, find the probability that she misses fewer than 3.

Odd one out: For $X \sim B(10, 0.3)$, three of the following expressions all equal $P(X \geq 3)$. Which one does NOT?

Revisit your thinking

Earlier you estimated $P(X \geq 1)$ for a fair coin tossed $5$ times and chose between direct sum and complement.

The complement is dramatically faster: $P(X \geq 1) = 1 - (0.5)^5 = \tfrac{31}{32}$ — one term instead of five. This pattern (right-tail probability → complement) is the single biggest time-saver in cumulative binomial questions.

auto-saved

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. $X \sim B(6, 0.4)$. Find $P(X \leq 1)$ to 4 decimal places. (2 marks)

auto-saved

ApplyBand 43 marks

Q2. A fair die is rolled $10$ times. Find the probability that a six appears at least twice. (3 marks)

auto-saved

AnalyseBand 53 marks

Q3. A factory ships batches of $25$ items. $8\%$ of items are defective. A batch is rejected if it contains more than 2 defective items. Find the probability that a batch is rejected, to 4 decimal places. (3 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:

1. $P(X=0) = (0.6)^8 \approx 0.0168$; $P(X=1) = 8(0.4)(0.6)^7 \approx 0.0896$; $P(X=2) = 28(0.4)^2(0.6)^6 \approx 0.2090$. Sum $\approx 0.3154$.

2. $P(X \geq 1) = 1 - (5/6)^{12} \approx 1 - 0.1122 = 0.8878$.

3. $X \sim B(15, 0.1)$. Complement (3 terms) beats direct (13 terms). $P(X=0)=(0.9)^{15}\approx 0.2059$; $P(X=1)=15(0.1)(0.9)^{14}\approx 0.3432$; $P(X=2)=105(0.01)(0.9)^{13}\approx 0.2669$. $P(X \leq 2) \approx 0.8160$, so $P(X > 2) \approx 0.1840$.

4. Both methods give $P(4 \leq X \leq 6) = (\binom{10}{4}+\binom{10}{5}+\binom{10}{6})/1024 = (210+252+210)/1024 = 672/1024 = 21/32 \approx 0.6563$.

5. $X$ = misses $\sim B(10, 0.3)$. $P(X=0)\approx 0.0282$; $P(X=1)\approx 0.1211$; $P(X=2)\approx 0.2335$. $P(X < 3) = P(X \leq 2) \approx 0.3828$.

Q1 (2 marks): $P(X=0) = (0.6)^6 = 0.0467$ [1]; $P(X=1) = 6(0.4)(0.6)^5 = 0.1866$. Sum $\approx 0.2333$ [1].

Q2 (3 marks): $X \sim B(10, 1/6)$ [1]. $P(X=0)=(5/6)^{10}\approx 0.1615$; $P(X=1)=10(1/6)(5/6)^9 \approx 0.3230$ [1]. $P(X \geq 2) = 1 - 0.1615 - 0.3230 \approx 0.5155$ [1].

Q3 (3 marks): $X \sim B(25, 0.08)$; want $P(X \geq 3) = 1 - P(X \leq 2)$ [1]. $P(X=0)=(0.92)^{25}\approx 0.1244$; $P(X=1)=25(0.08)(0.92)^{24}\approx 0.2704$; $P(X=2)=300(0.08)^2(0.92)^{23}\approx 0.2821$ [1]. $P(X \leq 2) \approx 0.6769$; $P(\text{reject}) \approx 0.3231$ [1].

Boss battle · The Cumulative Crusher

earn bronze · silver · gold

Five timed questions on cumulative binomial probabilities — left-tail, right-tail, and intervals. 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 cumulative-binomial questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 06 · Binomial Probability Lesson 08 · Mean of the Binomial Distribution →

Module overview · Extension 1 · Checkpoint 2