Binomial probabilities and cumulative probabilities: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

“Recognise binomial conditions and calculate exact and cumulative binomial probabilities, using complements for at-least and at-most events”

8 min readUpdated 2 Jun 2026

Quick answer

A binomial variable $X\sim\mathrm{B}(n,p)$ has $P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k}$ . Cumulative events combine terms; use the complement, for example $P(X\ge 1)=1-P(X=0)$ , to avoid long sums.

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What this dot point is asking
Recognising the binomial conditions
Exact probabilities
Cumulative probabilities
At-most and between events
Choosing the efficient route
The shape of the binomial coefficients
Exam-style questions

What this dot point is asking

SCSA Unit 3 asks you to recognise when the binomial model applies and to compute both exact and cumulative probabilities. This dot point appears in both sections; cumulative calculations are usually done with the calculator in Section Two but the reasoning is examined throughout.

Recognising the binomial conditions

Before using the formula, confirm the four conditions are met.

Binomial conditions

A fixed number of trials $n$ .
Each trial has two outcomes (success or failure).
The success probability $p$ is constant across trials.
The trials are independent.

If any condition fails, for example sampling without replacement from a small population (which breaks independence), the binomial model does not apply.

Exact probabilities

Binomial probability

For $X\sim\mathrm{B}(n,p)$ ,

P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k}, \qquad k=0,1,\dots,n,

where

\binom{n}{k}=\dfrac{n!}{k!(n-k)!}

counts the arrangements of

k

successes among

n

trials.

An exact binomial probability

An archer fires $5$ independent shots, each hitting with probability $p=0.8$ . We calculate the probability of exactly two hits.

Step 1: Identify the parameters

The variable $X$ counts hits in $n=5$ independent trials with constant success probability $p=0.8$ , so $X\sim\mathrm{B}(5,0.8)$ and we want $k=2$ . Confirming the parameters before substituting prevents errors with the exponents.

Step 2: Apply the binomial formula

Substitute $n=5$ , $k=2$ , $p=0.8$ and $1-p=0.2$ into the formula.

P(X=2)=\binom{5}{2}(0.8)^{2}(0.2)^{3}.

The binomial coefficient counts the number of ways to place exactly $2$ successes among $5$ trials, and the two power terms give the probability of each such arrangement.

Step 3: Evaluate each factor

Compute the three factors separately to avoid arithmetic mistakes.

\binom{5}{2}=10, \qquad (0.8)^{2}=0.64, \qquad (0.2)^{3}=0.008.

P(X=2)=10\times 0.64\times 0.008 = 0.0512.

Final answer: $P(X=2)=0.0512$ , so there is about a $5.1\%$ chance of exactly two hits. Markers reward the correct coefficient and the matching powers of $p$ and $1-p$ .

Cumulative probabilities

At-least and at-most questions sum several exact terms. The complement is the efficient route, especially for at-least-one events.

At least one success using the complement

At least one hit

For the archer above, find $P(X\ge 1)$ .

The complement of at least one hit is no hits.

P(X\ge 1)=1-P(X=0).

Compute $P(X=0)=\binom{5}{0}(0.8)^{0}(0.2)^{5}=(0.2)^{5}=0.00032$ .

P(X\ge 1)=1-0.00032 = 0.99968.

So a hit is almost certain. Using the complement replaces a five-term sum with a single calculation. In the calculator-assumed section the cumulative binomial function gives this directly.

At-most and between events

For at-most events, $P(X\le k)$ sums terms from $0$ to $k$ . For a range, $P(a\le X\le b)$ sums from $a$ to $b$ , or use $P(X\le b)-P(X\le a-1)$ with the cumulative function. Reading the inequality precisely is essential: $P(X\ge 2)$ excludes $X\le 1$ , so $P(X\ge 2)=1-P(X\le 1)$ .

Translating inequalities into cumulative terms

Let $X\sim\mathrm{B}(12,0.3)$ . We find $P(3\le X\le 6)$ using the cumulative binomial function available in the calculator-assumed section.

Step 1: Express the range as a difference of cumulative probabilities

Rewrite the range so it involves two cumulative ( $\le$ ) probabilities, which the calculator can evaluate directly.

P(3\le X\le 6)=P(X\le 6)-P(X\le 2).

Subtracting $P(X\le 2)$ removes the unwanted cases $X=0,1,2$ while keeping $X=3,4,5,6$ . The lower limit $3$ must be included, so you subtract $P(X\le 2)$ , not $P(X\le 3)$ .

Step 2: Evaluate using the calculator

Use the cumulative binomial function with $n=12$ , $p=0.3$ to obtain each value.

P(X\le 6)\approx 0.961, \qquad P(X\le 2)\approx 0.253.

Reading these values directly from the calculator is the efficient approach in Section Two; summing four exact terms would take far longer.

Step 3: Subtract and state the result

P(3\le X\le 6)\approx 0.961-0.253=0.708.

Final answer: $P(3\le X\le 6)\approx 0.708$ . Listing the included values before choosing the cumulative endpoints prevents off-by-one errors.

Choosing the efficient route

For at-least-one events the complement $1-P(X=0)$ is almost always fastest. For a small upper limit, summing exact terms directly may be quicker than the calculator menu. For a wide range, a difference of cumulative values is best. SCSA examiners accept any valid method, but a clear statement of which probabilities are being added or subtracted earns method marks even if an arithmetic slip occurs later.

The shape of the binomial coefficients

The coefficients $\binom{n}{k}$ are the entries of row $n$ of Pascal's triangle, and they are symmetric: $\binom{n}{k}=\binom{n}{n-k}$ . This is why, when $p=0.5$ , the distribution of $X$ is symmetric about $\dfrac{n}{2}$ , so $P(X=k)=P(X=n-k)$ . For example with $n=4$ and $p=0.5$ , the probabilities $0.0625, 0.25, 0.375, 0.25, 0.0625$ mirror each other. When $p<0.5$ the distribution is skewed toward the low values, and when $p>0.5$ toward the high values. Knowing the symmetry lets you check a calculator answer quickly and predict which tail carries more probability before computing anything.

Common errors

Applying the binomial when trials are dependent: Sampling without replacement from a small population breaks independence, so $p$ is not constant.
Swapping the powers: In $\binom{n}{k}p^{k}(1-p)^{n-k}$ , the exponent on $p$ is the number of successes $k$ and on $1-p$ is the number of failures $n-k$ .
Misreading the inequality: $P(X\ge 2)=1-P(X\le 1)$ , not $1-P(X\le 2)$ . List which values are included.
Summing every term unnecessarily: Use the complement for at-least-one events instead of adding many terms.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style7 marks

Calculator-assumed. In a large batch,

8\%

of components are defective. A sample of

20

components is selected. Let

X

be the number of defectives. (a) State the distribution of

X

. (b) Find

P(X=2)

. (c) Find

P(X\ge 3)

. (d) Find the probability that at least one but no more than three are defective.

Show worked answer →

A standard calculator-assumed binomial question with cumulative parts.

(a) $X\sim\mathrm{B}(20,0.08)$ : fixed $n=20$ , two outcomes, constant $p=0.08$ , independent trials.

(b) $P(X=2)=\binom{20}{2}(0.08)^{2}(0.92)^{18}=190\times 0.0064\times 0.2229\approx 0.271$ .

(c) Use the complement: $P(X\ge 3)=1-P(X\le 2)$ . With the cumulative binomial, $P(X\le 2)\approx 0.789$ , so $P(X\ge 3)\approx 0.211$ .

(d) $P(1\le X\le 3)=P(X\le 3)-P(X\le 0)\approx 0.929-0.189=0.740$ .

Markers reward the correctly stated distribution, the exact term in (b), use of the complement in (c), and the difference of cumulative values in (d).

WACE 2023-style4 marks

Calculator-free. A multiple-choice quiz has

6

questions, each with

4

options and exactly one correct answer. A student guesses every answer. (a) Find the probability of getting exactly

2

correct. (b) Find the probability of getting at least one correct.

Show worked answer →

A calculator-free binomial with friendly numbers ( $p=\tfrac{1}{4}$ ).

(a) $X\sim\mathrm{B}\!\left(6,\tfrac{1}{4}\right)$ . $P(X=2)=\binom{6}{2}\left(\tfrac{1}{4}\right)^{2}\left(\tfrac{3}{4}\right)^{4}=15\times\dfrac{1}{16}\times\dfrac{81}{256}=\dfrac{1215}{4096}\approx 0.297$ .

(b) Use the complement: $P(X\ge 1)=1-P(X=0)=1-\left(\tfrac{3}{4}\right)^{6}=1-\dfrac{729}{4096}=\dfrac{3367}{4096}\approx 0.822$ .

Markers reward the binomial coefficient, exact fractions, and the complement in (b) rather than summing six terms.

Binomial probabilities and cumulative probabilities: WACE Year 12 Mathematics Methods Unit 3

What this dot point is asking

Recognising the binomial conditions

Exact probabilities

Step 1: Identify the parameters

Step 2: Apply the binomial formula

Step 3: Evaluate each factor

Cumulative probabilities

At least one hit

At-most and between events

Step 1: Express the range as a difference of cumulative probabilities

Step 2: Evaluate using the calculator

Step 3: Subtract and state the result

Choosing the efficient route

The shape of the binomial coefficients

Exam-style questions

Practise this

Sources & how we know this