6.3 - Venn probabilities and mutually exclusive events

Learn to read probabilities from Venn regions, reconstruct missing counts and decide when event probabilities can be added. You will distinguish a whole circle from its “only” region and count each person once. This lesson applies at Foundation and Higher; its addition rule is for mutually exclusive events.

Reading Venn regions as events

In a group of $40$ learners, $20$ attend music and $18$ attend sport. Does that mean $38$ attend at least one club? What if $8$ attend both? The overlap is the key to this question.

You already know

An event selects outcomes from the universal set $\mathrm{E}$ . For a person chosen uniformly at random, probability is the selected count divided by the total population.

A Venn diagram gives each person one place. Someone in both clubs goes in the overlap, rather than being drawn twice. The circles show membership; their areas do not measure the probabilities.

Venn events

The intersection $A\cap B$ means both $A$ and $B$ . The union $A\cup B$ means $A$ , $B$ or both. The complement $A'$ means outside $A$ , but still inside $\mathrm{E}$ . “ $A$ only” excludes the overlap.

Read “or” inclusively unless the question says “exactly one”. In this diagram the four separate regions contain $12$ , $8$ , $10$ and $10$ learners, totalling $40$ .

[DIAGRAM: asset_slug: m094_venn_example; caption: Clubs of forty learners, with twelve music only, eight both, ten sport only and ten neither; description: Universal set E contains 40 learners. Music-only region 12, intersection 8, sport-only 10, outside both 10. Circles are schematic and their areas do not represent frequency; file: m094_venn_example.png; recommended_method: matplotlib]
Clubs of forty learners, with twelve music only, eight both, ten sport only and ten neither

The overlap contributes to either whole-circle total. To find the union, count that overlap once. The counting probability rule is absent from both tier formulae sheets.

Worked example 1: selecting Venn regions

Question. Of $40$ learners, $12$ attend music only, $8$ both clubs, $10$ sport only and $10$ neither. One learner is chosen at random. Find the probabilities of music, both, neither, and music or sport.

Plan. Select the separate regions for each event and divide by $40$ .

\begin{aligned} P(\text{music})&=\frac{12+8}{40}=\frac12 &&\text{whole music circle}\\ P(\text{both})&=\frac8{40}=\frac15 &&\text{overlap only}\\ P(\text{neither})&=\frac{10}{40}=\frac14 &&\text{outside both}\\ P(\text{music or sport})&=\frac{12+8+10}{40}=\frac34 &&\text{three inside regions} \end{aligned}

Why include $8$ in music? Those learners belong to music even though they also belong to sport.

Answer. Respectively $\frac12$ , $\frac15$ , $\frac14$ and $\frac34$ .

Check. The union and neither are complements: $\frac34+\frac14=1$ .

Change only the first target from music to sport. Keep the population and the other requested events unchanged.

Worked example 2: selecting the other circle

Question. Of $40$ learners, $12$ attend music only, $8$ both, $10$ sport only and $10$ neither. One is chosen at random. Find the probabilities of sport, both, neither, and music or sport.

Plan. Include the overlap in the sport circle and reuse the unchanged regions for the other events.

\begin{aligned} P(\text{sport})&=\frac{8+10}{40}=\frac9{20} &&\text{whole sport circle}\\ P(\text{both})&=\frac8{40}=\frac15 &&\text{same overlap}\\ P(\text{neither})&=\frac{10}{40}=\frac14 &&\text{same outside region}\\ P(\text{music or sport})&=\frac{30}{40}=\frac34 &&\text{same union} \end{aligned}

Why do three answers stay fixed? Changing which circle you ask about does not change the data.

Answer. $\frac9{20}$ , $\frac15$ , $\frac14$ and $\frac34$ .

Check. Sport has $18$ members and its complement has $12+10=22$ ; $18+22=40$ .

The opening puzzle gives a useful surprise: $38$ club memberships belong to only $30$ people. The $8$ people in both clubs each appear in both membership totals.

Watch out

$P(\text{music})=\frac{12}{40}$ answers “music only”. For music, include the overlap: $\frac{12+8}{40}=\frac12$ .

Use the same region choices with fresh counts in the next check. Every required count is supplied here.

Reconstructing region counts

A statement that $28$ people belong to $A$ normally gives the whole-circle total. It includes people who also belong to $B$ . The notation $n(A)$ means the number of members of $A$ .

Start with the overlap, the most restricted region. Subtract it from each circle total to find the only regions. Add the three inside regions, then subtract their total from the universal total to find neither. Each subtraction removes people already placed elsewhere.

Worked example 3: reconstructing the four regions

Question. Of $50$ people, $28$ belong to $A$ , $19$ to $B$ , and $7$ to both. Find all four region counts and the probability of exactly one membership for a randomly chosen person.

Plan. Place both first, find each only count, then find neither.

\begin{aligned} \text{Both}&=7 &&\text{given overlap}\\ A\text{ only}&=28-7=21 &&\text{remove both from }A\\ B\text{ only}&=19-7=12 &&\text{remove both from }B\\ \text{Neither}&=50-(21+7+12)=10 &&\text{remaining people}\\ P(\text{exactly one})&=\frac{21+12}{50}=\frac{33}{50} &&\text{exclude both} \end{aligned}

Why start with both? It belongs to both circle totals, so it must be removed from each to find the only regions.

Answer. $21$ in $A$ only, $7$ both, $12$ in $B$ only, $10$ neither; probability $\frac{33}{50}$ .

Check. $21+7=28$ , $12+7=19$ , and all four regions total $50$ .

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Number

Algebra

Shape, space and measure

Handling data

6.3 - Venn probabilities and mutually exclusive events

Reading Venn regions as events

Reconstructing region counts

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