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SAMath MethodsSyllabus dot point

How do we model the number of successes in a fixed number of independent trials?

The Bernoulli distribution models a single success/failure trial, and the binomial distribution counts successes across n independent trials with constant probability.

When to use the binomial distribution, its probability formula, the mean and variance results, and how to recognise the four binomial conditions.

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  1. What this dot point is asking
  2. The four binomial conditions
  3. The binomial probability formula
  4. "At least" and "at most" probabilities
  5. Mean and variance
  6. Common errors
  7. Why it matters

What this dot point is asking

A Bernoulli trial is a single experiment with exactly two outcomes, labelled success and failure. The binomial distribution arises when you repeat such a trial a fixed number of times and count how many successes occur.

The four binomial conditions

You may only use the binomial model when all four hold:

  1. A fixed number nn of trials.
  2. Each trial has only two outcomes (success/failure).
  3. Trials are independent.
  4. The probability of success pp is constant across trials.

The binomial probability formula

If XB(n,p)X\sim B(n,p), the probability of exactly xx successes is:

The three pieces are: (nx)\binom{n}{x} (how many ways to choose which trials succeed), pxp^x (those successes), and (1p)nx(1-p)^{n-x} (the remaining failures).

"At least" and "at most" probabilities

For cumulative questions, add the relevant terms - and use the complement when it is shorter.

Mean and variance

The binomial mean and variance follow directly from the parameters:

For the quiz, E(X)=8×0.25=2E(X)=8\times 0.25=2 correct on average, and Var(X)=8×0.25×0.75=1.5\operatorname{Var}(X)=8\times 0.25\times 0.75=1.5.

Common errors

Why it matters

The binomial distribution is the most heavily examined discrete model in Stage 2 and is the bridge to the normal distribution in Topic 5, where a binomial with large nn is approximated by a normal curve. Recognising the four conditions also trains the modelling judgement assessed in the Mathematical Investigation.

Exam-style practice questions

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

2023 SACE Stage 24 marksGold cards occur by chance in packs of 10 trading cards, with the probability that a single card is gold being 0.035. Modelling the number of gold cards in a pack with a binomial distribution, calculate: (1) the expected number of gold cards; (2) the probability of exactly one gold card; (3) the probability of more than two gold cards.
Show worked answer →

Here X is binomial with n = 10 and p = 0.035.

(1) Expected number = np = 10 times 0.035 = 0.35 gold cards. (1 mark)

(2) P(X = 1) = C(10, 1) (0.035)^1 (0.965)^9 = 10 times 0.035 times 0.7238 = 0.253 (to three significant figures). (1 mark)

(3) P(X > 2) = 1 - P(0) - P(1) - P(2).
P(0) = (0.965)^10 = 0.6983, P(1) = 0.253, P(2) = C(10, 2)(0.035)^2(0.965)^8 = 45 times 0.001225 times 0.7501 = 0.0413.
So P(X > 2) = 1 - 0.6983 - 0.253 - 0.0413 = 0.00686 (to three significant figures). (2 marks)

Markers reward using the binomial formula with the correct n and p, and computing "more than two" as the complement of P(0) + P(1) + P(2).

2017 SACE Stage 22 marksA seed distributor claims 60% of seeds are viable. Let X be the number of viable seeds in a planting of 8 seeds, modelled by a binomial distribution. (i) State the expected number of viable seeds. (ii) State the most likely number of viable seeds.
Show worked answer →

X is binomial with n = 8 and p = 0.6.

(i) Expected number = np = 8 times 0.6 = 4.8 viable seeds. (1 mark)

(ii) The most likely value is the mode - the value of k with the largest P(X = k). Comparing the two values nearest the mean: P(X = 4) = C(8, 4)(0.6)^4(0.4)^4 = 0.232, and P(X = 5) = C(8, 5)(0.6)^5(0.4)^3 = 0.279. Since P(X = 5) is the larger, the most likely number of viable seeds is 5. (1 mark)

Note the mode (5) differs from the mean (4.8): the expected value need not be a whole number, whereas the most likely outcome must be. Markers expect the mode identified as the value with the greatest binomial probability.