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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.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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

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What this dot point is asking
The four binomial conditions
The binomial probability formula
"At least" and "at most" probabilities
Mean and variance
The shape of the binomial distribution
Common errors
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:

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

The binomial probability formula

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

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

Exactly three successes

A multiple-choice quiz has $8$ questions, each with $4$ options. A student guesses every answer. Find the probability of getting exactly $3$ correct.

Step 1: Identify the binomial parameters

Each question is an independent trial with two outcomes: correct (success) or incorrect (failure). There are $n = 8$ trials, and since each question has 4 options and one correct answer, the probability of success on any one trial is $p = \tfrac{1}{4} = 0.25$ . So $X \sim B(8, 0.25)$ counts the number correct.

Step 2: Write the binomial probability formula

The formula for exactly $x$ successes in $n$ trials is:

P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}.

Here $x = 3$ , $n = 8$ , $p = 0.25$ , and $1 - p = 0.75$ . Substituting:

P(X = 3) = \binom{8}{3}(0.25)^3(0.75)^{5}.

Step 3: Evaluate each component

The binomial coefficient $\binom{8}{3}$ counts the number of ways to choose which 3 of the 8 questions are correct.

\binom{8}{3} = \frac{8!}{3!\,5!} = 56.

The probability factors are $(0.25)^3 = 0.015625$ and $(0.75)^5 \approx 0.237305$ .

Step 4: Multiply to get the final probability

P(X = 3) = 56 \times 0.015625 \times 0.237305 \approx 0.2076.

Final answer: There is about a $20.8\%$ chance of exactly three correct.

"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, $E(X)=8\times 0.25=2$ correct on average, and $\operatorname{Var}(X)=8\times 0.25\times 0.75=1.5$ , giving a standard deviation of $\sqrt{1.5}\approx 1.22$ .

The shape of the binomial distribution

The value of $p$ controls the shape. When $p=0.5$ the distribution is symmetric about the mean $np$ . When $p<0.5$ it is skewed to the right (a long tail of higher values), and when $p>0.5$ it is skewed to the left. As $n$ grows large, the binomial distribution becomes increasingly bell-shaped, which is exactly why a binomial with large $n$ can be approximated by the normal distribution you meet in Topic 5. Recognising this link helps you sanity-check answers: the most probable outcome (the mode) always sits at or next to $np$ , so a computed probability that peaks far from $np$ signals an arithmetic error.

Finding n from a probability condition

An archer hits the bullseye with probability $0.3$ on each independent shot. How many shots are needed so that the probability of at least one bullseye exceeds $0.9$ ?

Step 1: Use the complement to write the probability

Calculating "at least one bullseye" directly requires adding $P(X=1) + P(X=2) + \cdots$ , which is tedious. It is far easier to use the complement: "at least one" is the opposite of "none at all."

P(\text{at least one}) = 1 - P(\text{none}) = 1 - (0.7)^n,

since each shot independently misses with probability $1 - 0.3 = 0.7$ , and all $n$ shots must miss for the "none" event.

Step 2: Set up and simplify the inequality

We need $P(\text{at least one}) > 0.9$ , so:

1 - (0.7)^n > 0.9 \quad\Rightarrow\quad (0.7)^n < 0.1.

Step 3: Solve for $n$ using logarithms

Taking the natural logarithm of both sides turns the exponential inequality into a linear one.

n \ln(0.7) < \ln(0.1).

Because $\ln(0.7) < 0$ , dividing both sides by a negative number reverses the inequality.

n > \frac{\ln(0.1)}{\ln(0.7)} \approx \frac{-2.3026}{-0.3567} \approx 6.46.

Step 4: Round up to the nearest whole number

Since $n$ must be a whole number of shots and $n > 6.46$ , the smallest valid value is $n = 7$ .

Final answer: $n = 7$ shots are needed.

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 $n$ 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.

SACE 20234 marksCalculator-assumed. Gold 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 (a) the expected number of gold cards, (b) the probability of exactly one gold card, and (c) the probability of more than two gold cards.

Show worked answer →

Here $X \sim B(10, 0.035)$ .

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

(b) $P(X = 1) = \binom{10}{1}(0.035)^1(0.965)^9 = 10 \times 0.035 \times 0.7238 \approx 0.253$ . (1 mark)

(c) $P(X > 2) = 1 - P(0) - P(1) - P(2)$ where $P(0) = (0.965)^{10} \approx 0.6983$ , $P(1) \approx 0.253$ , and $P(2) = \binom{10}{2}(0.035)^2(0.965)^8 \approx 0.0413$ . So $P(X > 2) \approx 1 - 0.6983 - 0.253 - 0.0413 \approx 0.00686$ . (2 marks)

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

SACE 20222 marksCalculator-assumed. A 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. (a) State the expected number of viable seeds. (b) State the most likely number of viable seeds.

Show worked answer →

$X \sim B(8, 0.6)$ .

(a) Expected number $= np = 8 \times 0.6 = 4.8$ viable seeds. (1 mark)

(b) The most likely value is the mode. Comparing the values nearest the mean: $P(X = 4) = \binom{8}{4}(0.6)^4(0.4)^4 \approx 0.232$ and $P(X = 5) = \binom{8}{5}(0.6)^5(0.4)^3 \approx 0.279$ . Since $P(X = 5)$ is larger, the most likely number 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.

SACE 20213 marksCalculator-assumed. A test has 12 multiple-choice questions, each with 5 options. A student guesses every answer. Let

X

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

X

. (b) Find

P(X \ge 3)

Show worked answer →

(a) Each guess succeeds with $p = \tfrac{1}{5} = 0.2$ , so $X \sim B(12, 0.2)$ . (1 mark)

(b) Use the complement: $P(X \ge 3) = 1 - P(0) - P(1) - P(2)$ . With $P(0) = (0.8)^{12} \approx 0.0687$ , $P(1) = \binom{12}{1}(0.2)(0.8)^{11} \approx 0.2062$ , and $P(2) = \binom{12}{2}(0.2)^2(0.8)^{10} \approx 0.2835$ ,

P(X \ge 3) \approx 1 - 0.0687 - 0.2062 - 0.2835 \approx 0.442.

Marks: one for stating $B(12, 0.2)$ , one for the complement set-up, one for the value. (2 marks for part b.)