CCEA CCEA Unit 3 (style)-style practice: A fair die is rolled 5 times. Find the probability of getting exactly two sixes.

This is binomial with n = 5 trials, success "a six" with p = 16, and r = 2. Use P(X = r) = nrp^r(1-p)^n-r: P(X = 2) = 52(16)^2(56)^3. Compute: 52 = 10, so P(X = 2) = 10 × 136 × 125216 = 12507776 = 0.161 to 3 significant figures. Marks are for identifying n, p and r, the formula, and the value. Forgetting the 52 coefficient is the usual error.

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How do you use the binomial distribution to find probabilities for a fixed number of independent trials?

Use the binomial distribution: identify when it applies, calculate probabilities of r successes in n trials, and find its mean and variance.

A CCEA GCSE Further Mathematics answer on the binomial distribution, covering the conditions for its use, the probability formula for r successes in n trials, and the mean and variance in the Statistics unit.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
When the binomial applies
The probability formula
At least and at most
Mean and variance
Why this matters

What this dot point is asking

The binomial distribution models the number of successes in a fixed number of independent trials, and CCEA GCSE Further Mathematics expects you to use it confidently. You must recognise when a situation is binomial, calculate the probability of exactly $r$ successes in $n$ trials, handle "at least" and "at most" using the complement, and state the mean and variance. This is the first named distribution of the Statistics unit and builds on the multiplication law for independent events.

When the binomial applies

The binomial distribution has four defining conditions, and a question is only binomial if all four hold. Checking them is the first step.

Coin tosses, dice rolls for a particular face, and sampling with replacement all fit; sampling without replacement does not, because the probability changes between trials.

The probability formula

The formula combines two parts: the chance of one particular arrangement of $r$ successes and $n - r$ failures, and the number of different arrangements.

The $p^r$ accounts for the $r$ successes and the $(1-p)^{n-r}$ for the failures; the coefficient $\binom{n}{r}$ multiplies up because the successes can occur in many orders.

At least and at most

Questions often ask for a range of values rather than an exact count. "At most" means add the probabilities up to that value, while "at least one" is far quicker through the complement.

At least two successes

A multiple-choice quiz has $6$ questions, each guessed with probability $0.25$ of being right. Find the probability of getting at least two correct.

step 1 Set up the distribution

This is $X \sim B(6, 0.25)$ , and "at least two" means $P(X \ge 2)$ .

step 2 Use the complement

$P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$ , which is shorter than adding five terms.

step 3 Compute the two small terms

$P(X = 0) = (0.75)^6 = 0.1780$ . $P(X = 1) = \binom{6}{1}(0.25)(0.75)^5 = 6 \times 0.25 \times 0.2373 = 0.3560$ .

step 4 Subtract

$P(X \ge 2) = 1 - 0.1780 - 0.3560 = 0.466$ to 3 significant figures.

Mean and variance

For a binomial distribution the average number of successes and its spread have simple formulae. The mean is $np$ , which matches intuition: with probability $p$ on each of $n$ trials you expect $np$ successes. The variance is $np(1-p)$ , and its square root is the standard deviation. These let you describe the distribution without listing every probability, and they connect the binomial to the wider study of distributions in the unit.

Why this matters

The binomial distribution is the workhorse model for counting successes and a cornerstone of the Statistics unit. It applies the multiplication law for independent events in a systematic way, its complement technique trains efficient problem solving, and its mean and variance link to the measures of location and spread studied elsewhere. For large $n$ it also connects to the normal distribution, which approximates it, so a firm grasp here supports the later distribution work.

Exam-style practice questions

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

CCEA Unit 3 (style)4 marksA fair die is rolled

5

times. Find the probability of getting exactly two sixes.

Show worked answer →

This is binomial with $n = 5$ trials, success "a six" with $p = \dfrac{1}{6}$ , and $r = 2$ .

Use $P(X = r) = \binom{n}{r}p^r(1-p)^{n-r}$ : $P(X = 2) = \binom{5}{2}\left(\dfrac{1}{6}\right)^2\left(\dfrac{5}{6}\right)^3$ .

Compute: $\binom{5}{2} = 10$ , so $P(X = 2) = 10 \times \dfrac{1}{36} \times \dfrac{125}{216} = \dfrac{1250}{7776} = 0.161$ to 3 significant figures.

Marks are for identifying $n$ , $p$ and $r$ , the formula, and the value. Forgetting the $\binom{5}{2}$ coefficient is the usual error.

CCEA Unit 3 (style)4 marks

20\%

of components are faulty. In a sample of

8

, find the probability that at least one is faulty, and state the expected number of faulty components.

Show worked answer →

Binomial with $n = 8$ , $p = 0.2$ . "At least one" is easiest via the complement: $P(X \ge 1) = 1 - P(X = 0)$ .

$P(X = 0) = \binom{8}{0}(0.2)^0(0.8)^8 = 0.8^8 = 0.1678$ , so $P(X \ge 1) = 1 - 0.1678 = 0.832$ .

Expected number (mean) is $np = 8 \times 0.2 = 1.6$ .

Marks are for the complement method, the value, and the mean. Trying to add many separate terms instead of using the complement wastes time and invites errors.

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Sources & how we know this

CCEA GCSE Further Mathematics specification (2330) — CCEA (2017)