2.04b, 2.04c - Binomial models and probabilities | OCR A-Level Mathematics A

Recognise when a binomial model fits a count and explain its assumptions in context (2.04b). Calculate exact, cumulative and interval probabilities with the formula or a calculator, keeping the event's integer boundaries clear (2.04c).

Count successes in a fixed set of trials

How likely are at least two failed components in a pack of six? Suppose each component fails inspection with probability 0.25, independently of the others. Counting all the possible pass–fail lists by hand would mean handling 64 lists. A binomial model gathers them by their failure counts.

You already know

Independent events have product probabilities, and disjoint alternatives have additive probabilities. The combination $\binom nr$ counts choices of $r$ positions from $n$ ; order within that choice does not matter.

Call a failure a “success” for the count: success is a label, not praise for the manufacturer.

Binomial model

A binomial random variable counts successes in a fixed number $n$ of independent trials. Each trial has exactly two exhaustive outcomes, success and failure, and the same success probability $p$ . Write $X\sim\mathrm B(n,p)$ , with $q=1-p$ . Its possible counts are $0,1,\ldots,n$ .

For the pack, define $X$ as the number failing inspection. Then $X\sim\mathrm B(6,0.25)$ . The pack size and pass–fail classification are fixed by the procedure. Common probability and independence are assumptions about the process, rather than consequences of counting six components.

Feature	What it means here	A reason to question it
Fixed trial number	Inspect all six components	Stopping at the first failure changes the procedure
Two outcomes	Every component passes or fails the stated inspection	An unresolved result needs a classification rule
Common $p$	Each component has failure probability 0.25	Different production lines may have different failure rates
Independence	One result does not change another's probability	A shared faulty setting may link failures

Sampling without replacement from a small finite batch is generally not exactly binomial: the proportion of failed items remaining changes. A large population may make that change negligible, but the approximation needs a contextual argument. Simply knowing a long-run failure percentage does not establish independence.

The census large data set gives another concrete distinction. Imagine choosing ten of its 348 aligned local authorities without replacement and counting those whose 2011 percentage aged 65 or over exceeds 20%. That threshold defines two outcomes, but the success proportion among the remaining authorities changes as you select. The count is generally not exactly binomial. The trial is an authority selection, rather than a selection of an individual resident.

Key idea

Define the count first, then separate the conditions established by the procedure from the assumptions about its chance behaviour.

A useful critique names the real process that could break a particular assumption.

Count the arrangements of a fixed number of successes

Exactly two failures can appear in 15 different positions among six components. Missing that factor counts just one pass–fail list.

Binomial probability

For $X\sim\mathrm B(n,p)$ , $q=1-p$ , and an integer $0\leq r\leq n$ ,

\mathrm P(X=r)=\binom nr p^rq^{n-r}

In the formula sheet.

Independence gives one specified list probability $p^rq^{n-r}$ . There are $\binom nr$ such disjoint lists, all with the same probability because $p$ is constant. This also links the distribution to the expansion of $(q+p)^n=1$ : its terms are precisely the count probabilities.

Worked example 1: exactly two failures

Question. For $X\sim\mathrm B(6,0.25)$ , calculate $\mathrm P(X=2)$ to three significant figures.

Plan. Multiply the probability of one two-failure list by its number of arrangements.

\begin{aligned} \binom62&=15 && \text{choose failure positions} \\ \mathrm P(X=2)&=15(0.25)^2(0.75)^4 && \text{two failures, four passes} \\ &=0.296630859375 && \text{evaluate without rounding} \end{aligned}

Why this step? FFPPPP is only one of the 15 lists giving two failures.

Answer. $\mathrm P(X=2)=0.297$ , to three significant figures.

Check. One specified list has probability 0.019775390625; multiplying by 15 gives the result.

Change only the requested count, from two to three. The trial number and failure probability stay fixed.

Worked example 2: exactly three failures

Question. For the same model, calculate $\mathrm P(X=3)$ to three significant figures.

Plan. Choose three failure positions and match both exponents to the new count.

\begin{aligned} \binom63&=20 && \text{choose three positions} \\ \mathrm P(X=3)&=20(0.25)^3(0.75)^3 && \text{three of each outcome} \\ &=0.1318359375 && \text{evaluate} \end{aligned}

Why this step? Increasing the failure count also decreases the pass count; the exponents must still total six.

Answer. $\mathrm P(X=3)=0.132$ , to three significant figures.

Check. It is less than the probability of two failures: failures are relatively unlikely in this model.

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2.04b, 2.04c - Binomial models and probabilities

Count successes in a fixed set of trials

Count the arrangements of a fixed number of successes

Continue this lesson.

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