Statistics 3.2 - Conditional probability with diagrams and tables | Edexcel A-Level Mathematics

“Given that” changes which outcomes you consider. You will calculate conditional probabilities using event totals, Venn diagrams, trees and two-way tables, then choose a representation that makes the denominator clear.

Given that changes the sample space

A factory uses setting $H$ for $30\%$ of its components. Fault rates are $8\%$ with $H$ and $3\%$ without it. If a component is faulty, is the chance that $H$ was used still $30\%$ ? We will answer this by restricting the sample space.

You already know

$A\cap B$ means both events, and $A'$ means not $A$ . Individual Venn regions add to the whole event total.

Start with counts: of $100$ people, $60$ are in $B$ and $24$ are in both $A$ and $B$ . Given $B$ , only $60$ people remain eligible. The proportion also in $A$ is $24/60$ .

Conditional probability

$P(A\mid B)$ is the probability of $A$ given that $B$ occurs. It is defined when $P(B)>0$ . The condition after the vertical line restricts the sample space.

Divide the overlap by the whole condition.

Conditional probability

P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)>0.

Not in the formula booklet: learn it.

The booklet prints the equivalent intersection-product relation and a two-case expression for reversed conditioning. Learn this direct definition to identify the numerator and denominator.

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$Restrict to B, then find the fraction also in A$

Worked example 1: two directions

Question. Given $P(A)=0.48$ , $P(B)=0.60$ and $P(A\cap B)=0.24$ , find $P(A\mid B)$ , $P(B\mid A)$ and $P(A'\mid B)$ .

Plan. Divide by each condition total, using the remaining part of $B$ for the third probability.

\begin{aligned} P(A\mid B)&=0.24/0.60=0.40 &&\text{condition }B\\ P(B\mid A)&=0.24/0.48=0.50 &&\text{condition }A\\ P(A'\mid B)&=(0.60-0.24)/0.60=0.60 &&\text{remaining part of }B \end{aligned}

Why do the first two answers differ? The overlap is the same, but the restricted groups differ.

Answer. The probabilities are $0.40$ , $0.50$ and $0.60$ .

Check. Within $B$ , $A$ and $A'$ partition the possibilities: $0.40+0.60=1$ .

Change only $P(A)$ to $0.40$ . Only the question conditioned on $A$ needs a new denominator.

Worked example 2: a changed condition total

Question. Given $P(A)=0.40$ , $P(B)=0.60$ and $P(A\cap B)=0.24$ , find the same three probabilities.

Plan. Identify the condition separately for each fraction.

\begin{aligned} P(A\mid B)&=0.24/0.60=0.40 &&\text{unchanged condition}\\ P(B\mid A)&=0.24/0.40=0.60 &&\text{changed condition}\\ P(A'\mid B)&=0.36/0.60=0.60 &&\text{complement within }B \end{aligned}

Why not change every answer? Neither $B$ nor the overlap has changed.

Answer. The probabilities are $0.40$ , $0.60$ and $0.60$ .

Check. The separate regions $0.16$ , $0.24$ , $0.36$ and $0.24$ sum to $1$ .

The event after the line matters more than the order in which the data were presented.

Watch out

$P(A\mid B)$ generally differs from $P(B\mid A)$ . Read the condition before choosing the denominator.

Change the $A$ total again.

Using complements and unions first

Conditional questions may hide the numerator. Find the relevant region first: “neither $A$ nor $B$ ” is the complement of their union.

Complements, unions and intersections

For events $A$ and $B$ ,

\begin{aligned} P(A')&=1-P(A)\\ P(A\cup B)&=P(A)+P(B)-P(A\cap B)\\ P(A\cap B)&=P(A)P(B\mid A),\qquad P(A)>0 \end{aligned}

In the formula booklet.

The product relation follows by rearranging the conditional definition. Independence is unnecessary because the second probability is conditional.

Worked example 3: neither within a complement

Question. Given $P(A)=0.42$ , $P(B)=0.50$ and $P(A\cap B)=0.18$ , find $P(A'\mid B')$ .

Plan. Take the complement of the union, then divide by $P(B')$ .

\begin{aligned} P(A\cup B)&=0.42+0.50-0.18=0.74 &&\text{addition rule}\\ P(A'\cap B')&=1-0.74=0.26 &&\text{neither event}\\ P(A'\mid B')&=\frac{0.26}{1-0.50}=0.52 &&\text{restrict to }B' \end{aligned}

Why use the union's complement? The numerator must be outside both events.

Answer. The probability is $0.52$ .

Check. Within $B'$ , the $A$ -only region $0.24$ and the neither region $0.26$ sum to $0.50$ .

Change only $P(A)$ to $0.32$ . The denominator stays fixed while the outside-both numerator grows.

Worked example 4: a changed numerator

Question. Given $P(A)=0.32$ , $P(B)=0.50$ and $P(A\cap B)=0.18$ , find $P(A'\mid B')$ .

Plan. Recover neither from the union before conditioning.

\begin{aligned} P(A'\cap B')&=1-(0.32+0.50-0.18)=0.36 &&\text{outside both}\\ P(A'\mid B')&=0.36/0.50=0.72 &&\text{condition }B' \end{aligned}

Why does the answer rise? More of the unchanged group $B'$ lies outside $A$ .

Answer. The probability is $0.72$ .

Check. $P(A\mid B')=(0.32-0.18)/0.50=0.28$ , and $0.28+0.72=1$ .

Complementing the wanted event keeps the condition fixed.

Watch out

$P(A'\mid B')\ne1-P(A\mid B)$ in general. The valid rule is $P(A'\mid B')=1-P(A\mid B')$ : both fractions condition on $B'$ .

Find neither within the same condition using a new $A$ total.

Reading conditional probability from Venn diagrams

A Venn diagram helps with overlapping sets in one population. Fill the overlap first, subtract it from each event total, then find neither. Counts and probabilities both work, provided numerator and denominator use the same units.

Worked example 5: overlapping student groups

Question. Of $120$ students, $52$ study music, $65$ play sport and $25$ do both. Find $P(M\mid S)$ , $P(M'\cap S')$ and $P(M\cup S)$ .

Plan. Build the four separate regions, then select each denominator.

\begin{aligned} \text{music only}&=52-25=27 &&\text{remove overlap}\\ \text{sport only}&=65-25=40 &&\text{remove overlap}\\ \text{neither}&=120-(27+25+40)=28 &&\text{remaining students}\\ P(M\mid S)&=25/65=5/13 &&\text{sport denominator}\\ P(M'\cap S')&=28/120=7/30 &&\text{whole population}\\ P(M\cup S)&=92/120=23/30 &&\text{three inside regions} \end{aligned}

Why is the first denominator $65$ ? Knowing the student plays sport removes everybody outside that group.

Answer. The probabilities are $5/13$ , $7/30$ and $23/30$ .

Check. The four regions sum to $120$ , and the last two probabilities sum to $1$ .

Change only the overlap from $25$ to $20$ , keeping both group totals fixed.

Worked example 6: a changed overlap count

Question. Of $120$ students, $52$ study music, $65$ play sport and $20$ do both. Find the same three probabilities.

Plan. Rebuild the regions before forming the fractions.

\begin{aligned} \text{music only}&=32,\quad \text{sport only}=45 &&\text{subtract }20\\ \text{neither}&=120-(32+20+45)=23 &&\text{complete the space}\\ P(M\mid S)&=20/65=4/13 &&\text{condition sport}\\ P(M'\cap S')&=23/120 &&\text{unconditional}\\ P(M\cup S)&=97/120 &&\text{union count} \end{aligned}

Why does the union grow? Less of the fixed event totals is counted twice.

Answer. The probabilities are $4/13$ , $23/120$ and $97/120$ .

Check. $32+20+45+23=120$ , with no negative region.

A Venn sketch need not be to scale: use its labels rather than measuring its areas unless an area model is explicitly stated.

Exam tip

The conditional denominator includes every region inside the condition, including the overlap.

Change the overlap count again.

Using tree diagrams for staged events

Use a tree for stages. Each second-stage branch is conditional on the first-stage outcome. Branches need not match, and each set leaving one branch point must sum to $1$ .

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Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Statistics 3.2 - Conditional probability with diagrams and tables

Given that changes the sample space

Using complements and unions first

Reading conditional probability from Venn diagrams

Using tree diagrams for staged events

Continue this lesson.

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