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How do we calculate exact and cumulative binomial probabilities, including at-least and at-most events?

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

WACE Year 12 Mathematics Methods Unit 3 binomial probabilities: recognising the four conditions, computing exact probabilities, cumulative at-least and at-most events using the complement, with worked SCSA-style examples.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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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

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.

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

Exact probabilities

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-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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksCalculator-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 20234 marksCalculator-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.