M2 - Conditional probability | AQA A-Level Mathematics

Calculate probabilities after information restricts the possible outcomes. Use two-way tables, Venn diagrams and probability trees to identify the right conditioning population, including when a question reverses the information you were given.

Conditioning and tables

A production line makes $60\%$ of a factory's components. Must it also make most of the defective components? Not if its defect rate is sufficiently low. To answer, we need to be precise about which population a probability describes.

Start with a smaller table. Among $100$ trainees, $60$ attended training and $36$ of those passed. Once you are told the selected trainee attended training, only those $60$ people remain eligible. The probability of passing is $36/60$ , rather than $36/100$ .

You already know

An intersection describes both events occurring. For uniform selection from a finite population, an event's probability is its count divided by the eligible population count.

The information after the vertical bar determines that eligible population.

Conditional probability

$\mathrm P(A\mid B)$ is the probability of $A$ , given $B$ . Restrict attention to $B$ ; the successful part of this restricted population is $A\cap B$ .

Rescale the probability of $B$ to become the whole restricted population. Provided $\mathrm P(B)>0$ , this gives

\mathrm P(A\mid B)=\frac{\mathrm P(A\cap B)}{\mathrm P(B)}

The booklet supplies $\mathrm P(A\cap B)=\mathrm P(B)\mathrm P(A\mid B)$ , with event names interchanged from its printed version. The conditional ratio is its rearrangement; it is not separately printed. Learn how to rearrange it and the positive-denominator condition. A zero numerator with a positive denominator gives probability zero; a zero denominator does not define this elementary conditional probability.

Worked example 1: conditioning on a table row

Question. One trainee is selected uniformly from this table. Find $\mathrm P(P\mid T)$ , where $T$ means trained and $P$ means passed.

	Passed $P$	Did not pass $P'$	Total
Trained $T$	$36$	$24$	$60$
Not trained $T'$	$16$	$24$	$40$
Total	$52$	$48$	$100$

Plan. Use the trained row as the denominator and its passed cell as the numerator.

\begin{aligned} n(T)&=60 && \text{eligible trainees} \\ n(P\cap T)&=36 && \text{eligible successes} \\ \mathrm P(P\mid T)&=\frac{36}{60}=0.6 && \text{restricted proportion} \end{aligned}

Why this step? The condition tells us training has occurred; the $40$ untrained people do not enter this denominator.

Answer. $0.6$ .

Check. Of the trained row, the failing proportion is $24/60=0.4$ ; the two alternatives sum to $1$ .

Change only the condition from trained to not trained, keeping the same table and target event.

Worked example 2: conditioning on the other row

Question. For the same trainee table, with trained counts $36$ passed and $24$ failed, and untrained counts $16$ passed and $24$ failed, find $\mathrm P(P\mid T')$ .

Plan. Restrict to the untrained row.

\begin{aligned} n(T')&=16+24=40 && \text{new eligible total} \\ \mathrm P(P\mid T')&=\frac{16}{40}=0.4 && \text{untrained pass proportion} \end{aligned}

Why this step? Changing the condition changes both the selected row and the relevant pass count.

Answer. $0.4$ .

Check. $0.6(0.6)+0.4(0.4)=0.52$ , reproducing the whole-table pass proportion $52/100$ .

Reversing the bar changes the population in a different way: $\mathrm P(T\mid P)=36/52=9/13$ , because now every passer is eligible. The shared numerator does not make the two conditional probabilities equal. The table shows association; it does not by itself establish that training caused the difference.

Watch out

Wrong: $\mathrm P(P\mid T)=36/100$ . That is the joint probability of trained and passed. The conditional denominator is $60$ , giving $36/60$ .

If $A$ and $B$ are independent and $\mathrm P(B)>0$ , the ratio gives $\mathrm P(A\mid B)=\mathrm P(A)$ . Conversely, that equality gives the product condition. This conditional independence test is not separately printed in the booklet: learn the reasoning. It is not valid to divide by a probability-zero condition.

Conditioning in Venn diagrams

A Venn diagram makes the restricted population visible. For $\mathrm P(A\mid B')$ , keep every region outside $B$ , including the region outside all circles. The numerator keeps only the portion also in $A$ .

With three events, fill the centre first. Pairwise totals include that centre unless the question explicitly says “only”. Subtract the centre from each pair total; then subtract the filled overlap regions from each whole-event total. The final outside count is the population total minus the union. This is a construction from event counts, rather than an extra booklet formula.

Worked example 3: constructing a three-event diagram

Question. Of $90$ people, $45$ run $R$ , $40$ swim $S$ and $32$ cycle $C$ . The inclusive pairwise totals are $20$ for $R\cap S$ , $15$ for $R\cap C$ and $12$ for $S\cap C$ . Exactly $7$ do all three. Construct the region counts and find $\mathrm P(R\cap S\mid C')$ for uniform selection.

Plan. Fill the triple region, then pair-only and single-only regions; restrict the probability to non-cyclists.

\begin{aligned} n(R\cap S\cap C')&=20-7=13 && \text{run and swim only} \\ n(R\cap C\cap S')&=15-7=8 && \text{run and cycle only} \\ n(S\cap C\cap R')&=12-7=5 && \text{swim and cycle only} \\ n(\text{run only})&=45-13-8-7=17 \\ n(\text{swim only})&=40-13-5-7=15 \\ n(\text{cycle only})&=32-8-5-7=12 \end{aligned}

The interior total is $17+15+12+13+8+5+7=77$ , leaving $90-77=13$ outside.

Three-event Venn diagram for 90 people: run only 17, swim only 15, cycle only 12, pair-only counts 13, 8, 5, triple 7 and outside 13

\mathrm P(R\cap S\mid C')=\frac{13}{90-32}=\frac{13}{58}

Why this step? Only the run-and-swim region outside $C$ is successful; all non-cyclists, including those outside every circle, are eligible.

Answer. The diagram shows the eight counts; the probability is $\frac{13}{58}$ .

Check. The non-cyclist regions sum to $17+15+13+13=58$ ; each whole circle recreates its given total.

Change just the condition from non-cyclist to cyclist. The same completed diagram now selects a different denominator and overlap.

Worked example 4: a different conditioning region

Question. In the same $90$ -person diagram, $32$ cycle and $7$ do all three activities. Find $\mathrm P(R\cap S\mid C)$ .

Plan. Keep the whole cyclist circle, then select its part that also lies in both other circles.

\begin{aligned} n(C)&=32 && \text{eligible population} \\ n(R\cap S\cap C)&=7 && \text{triple region} \\ \mathrm P(R\cap S\mid C)&=\frac7{32} && \text{restricted ratio} \end{aligned}

Why this step? The $13$ people in $R\cap S\cap C'$ no longer satisfy the condition.

Answer. $\frac7{32}$ .

Check. The cyclist regions sum to $12+8+5+7=32$ , so the numerator is a subset of the denominator.

A union can also be the condition. Here $n(S\cup C)=40+32-12=60$ , while $n(R\cap(S\cup C))=13+8+7=28$ . Thus $\mathrm P(R\mid S\cup C)=28/60=7/15$ . Add each eligible region once; an outcome in all three still counts once.

Watch out

Wrong: “Run-and-swim only is $20$ .” The given inclusive total contains the $7$ cyclists in the centre. Subtract them: the pair-only count is $13$ .

The same selection of regions works when the diagram is specified by probabilities instead of counts.

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M2 - Conditional probability

Conditioning and tables

Conditioning in Venn diagrams

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