Binomial assumptions | Year 12

01 / LEARN

Understand the idea.

Four conditions must work together

A binomial count records successes in a fixed number $n$ of trials. Each trial must have two categories, the same success probability $p$ , and independence from the other trials. Identify these conditions before writing $X \sim Bin (n, p)$ .

A trial can have many raw outcomes while still being grouped into two categories. For example, classify a die roll as a success if it is $5$ or $6$ , and as a failure otherwise. The count concerns successes, not die scores.

Look closely at the process

Check how the trials are produced. Drawing without replacement from a small collection makes later probabilities depend on earlier draws. A changing process may violate constant $p$ , while shared conditions can make apparently separate trials dependent.

A small sampling fraction from a large population may support a binomial approximation, but this needs justification. A rule that stops at the first success fails the fixed- $n$ condition: the number of trials is itself random.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An appropriate grouping

A fair die is rolled independently 12 times. X counts scores of 5 or 6. State a model and justify it.

Worked solution

1
Define the counted event and fixed trial count.
$n = 12, success: score 5 or 6$
Each of the twelve rolls is classified as success or failure.
2
Calculate the common success probability.
$p = \frac{2}{6} = \frac{1}{3}$
Two of the six equally likely die scores are successes on every roll.
3
State the justified model.
$X \sim Bin (12, \frac{1}{3})$
The number of trials is fixed, rolls are independent, and the success probability is constant.

Answer

$X \sim Bin (12, 1 / 3)$ .

EXAMPLE 02

Why replacement matters

A bag has two red and three blue counters. Two are drawn without replacement. Explain why the red count is not exactly Bin(2, 0.4).

Worked solution

1
Compare second-draw probabilities after each first colour.
$\begin{aligned} P (R_{2} ∣ R_{1}) & = \frac{1}{4} \\ P (R_{2} ∣ B_{1}) & = \frac{2}{4} \end{aligned}$
Without replacement, the remaining composition depends on the first draw.
2
Find the actual probability of two red counters.
$P (R_{1} \cap R_{2}) = \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{10} = 0.10$
For this ordered pair, multiply the first probability by the relevant conditional probability.
3
Compare with the proposed binomial model.
$P_{Bin (2, 0.4)} (X = 2) = {(0.4)}^{2} = 0.16 \neq 0.10$
The draws are dependent, so the proposed independent-trial model is not exact.

Answer

Not exactly binomial. For example $P (two red) = (2 / 5) (1 / 4) = 0.1$ , not ${0.4}^{2} = 0.16$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

A fair coin is tossed until its first head. Is the number of tails before stopping binomial?

Show answer & reasoning

No.

No. Stopping at the first head allows different numbers of tosses, so there is no fixed number of trials for a binomial count.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

A fair die is rolled independently 8 times. $X$ counts even scores. State $n$ and $p$ .
2 marks
Ten bulbs are tested independently, each with fault probability 0.03. State a model for the number of faults.
2 marks
Four cards are drawn without replacement from a standard deck. Explain why the heart count is not exactly $Bin (4, \frac{1}{4})$ .
2 marks
A player takes 20 independent shots, with success probability 0.8 for the first 10 and 0.6 for the last 10. Why is the total not exactly $Bin (20, 0.7)$ ?
2 marks
An online poll stops when it has collected five positive responses. Identify the fixed- $n$ problem.
2 marks
A factory samples 10 items without replacement from 100,000 items. Explain a possible approximate binomial model and a remaining limitation.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A fair die is rolled independently 8 times. $X$ counts even scores. State $n$ and $p$ .
2 marks
There are three even scores out of six.
The stated independent, fixed rolls supply the other binomial conditions.
$n = 8, p = \frac{1}{2}$ .
Ten bulbs are tested independently, each with fault probability 0.03. State a model for the number of faults.
2 marks
Success means a fault.
There are ten fixed independent tests with common success probability 0.03.
$X \sim Bin (10, 0.03)$ .
Four cards are drawn without replacement from a standard deck. Explain why the heart count is not exactly $Bin (4, \frac{1}{4})$ .
2 marks
After drawing a heart there are 12 hearts among 51 cards.
After drawing a non-heart there are 13 hearts among 51; neither equals the unconditional first-draw probability.
The draws are dependent and the heart probability changes.
A player takes 20 independent shots, with success probability 0.8 for the first 10 and 0.6 for the last 10. Why is the total not exactly $Bin (20, 0.7)$ ?
2 marks
The mean probability being 0.7 does not make every trial have probability 0.7.
The total is a sum of two counts with different success probabilities.
The success probability is not constant across trials.
An online poll stops when it has collected five positive responses. Identify the fixed- $n$ problem.
2 marks
The stopping condition fixes successes, not trials.
A binomial model requires a predetermined number of trials.
The number of people polled is random.
A factory samples 10 items without replacement from 100,000 items. Explain a possible approximate binomial model and a remaining limitation.
3 marks
State $n = 10$ and estimate $p$ from appropriate process information.
The draws are still not exactly independent, and clustering or a changing production process could undermine the approximation.
The sample fraction is tiny, so the fault probability changes little; independence is approximate.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: discrete random variables and probability distributions, expected value and variance, transformations and independent sums, Bernoulli and binomial models. Original supporting lessons and exercises follow the approved Ascent chapter structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Four conditions must work together

Look closely at the process

See the method in action.

An appropriate grouping

Define the counted event and fixed trial count.

Calculate the common success probability.

State the justified model.

Why replacement matters

Compare second-draw probabilities after each first colour.

Find the actual probability of two red counters.

Compare with the proposed binomial model.

Try it yourself.

Check your understanding

Keep practising