On a Venn diagram, 15 own a watch and 7 also own a headset. How many own only a watch?

8. The 15 is the whole watch set, so it already includes the 7 who also own a headset. The watch-only region is the set total minus the overlap: 15 − 7 = 8. Check: 8 watch-only + 7 in the overlap = 15, the whole watch set. The trap answer 15 reads the label as the 'only' group, which double-counts the 7.

How do you write a probability from a Venn diagram?

Region ÷ universal set. The denominator is the whole group surveyed, not a region count. If 60 people are surveyed and 7 own both items, the probability a random person owns both is 7/60, not 7/39. The numerator includes the overlap when the set is asked for, and the denominator is always the universal set.

37 people are surveyed and 20 can be a trainer (3 of them do both jobs). What is P(trainer)?

20/37. The trainer set is 20 — the 17 trainer-only plus the 3 in the overlap — and we pick from all 37, so P(trainer) = 20/37. The traps are 17/37 (drops the 3 overlap from the trainer total), 20/32 (drops the 5 who can do neither from the whole), and 17/20 (uses the trainer set as the whole group). Region check: trainer-only 17 + reception-only 12 + both 3 + neither 5 = 37.

GCSE Maths Foundation · AQA · Probability

Reading Venn diagrams: the label is the whole set, the denominator is the universal set

Venn questions trip students who read the number against a set as the “only” group. Told 15 own a watch and 7 of them also own a headset, they answer 15 for watch-only — but the 15 is already the whole watch set, so watch-only is $15 - 7 = 8$ . The same belief puts a region count under a probability: $7/39$ instead of $7/60$ .

The thirty-second fix: the number against a set is the WHOLE set (it includes the overlap), so subtract the overlap to get the “only” part — and a probability is region ÷ universal set, so the denominator is the whole group, never a region.

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How to spot it in your own work

You read a set label as the “only” group — writing 15 for watch-only when $15 - 7 = 8$ takes out the overlap.
You put a region count on the bottom of a probability — writing $7/39$ instead of $7/60$ over the universal set.
You used the wrong subgroup as the denominator — dividing by the 120 total when the question fixed the group as the 80 children ( $30/80$ , not $30/120$ ).
You dropped the overlap from a set total — using $17/37$ for P(trainer) when the trainer set is $17 + 3 = 20$ , giving $20/37$ .

An exam question that triggers it

Here is the canonical AQA Foundation trigger (Nov24 P3 Q14 shape, calculator):

37 people are surveyed. 20 can work as a trainer (including 3 who can do both jobs), 15 can work in reception (including the same 3), and 5 can do neither. Work out the probability that a randomly chosen person can work as a trainer.

The misconception answer is $17/37$ — using the trainer-only count and forgetting the 3 in the overlap.

But the trainer set is $17 + 3 = 20$ , and we pick from all 37, so $P(\text{trainer}) = 20/37$ .

Why students fall for this

A number written against a set on a Venn diagram counts the whole set — the people with that feature, whether or not they also have another. So the overlap people are already inside that number. The student sees a single label and reads it as a single region, missing that the “only” part is the total minus the overlap.

The same belief about “the whole” corrupts the denominator of a probability. Asked for a chance, the student divides by whatever region is in view rather than the universal set — the total number of people surveyed. That is how $7/60$ becomes $7/39$ , and how a conditional asked “of the children” gets divided by the grand total instead of the children.

AQA Foundation calculator papers exploit every face of this: a region that includes the intersection (Nov24 P3 Q14b — reception-but-not-trainer is $15 - 3 = 12$ ), a wrong denominator (Jun23 P2 Q11b — $7/60$ not $7/39$ ), a conditional subgroup (Nov24 P1 Q7b — divide by the 80 children, not the 120 total), and criticising a Venn that ignores the overlap (Jun22 P1 Q20).

The fix: A label is the whole set: subtract the overlap for a region, and divide by the universal set for a probability

A labelled number is the whole set. It includes the overlap, so the “only” part is the set total minus the overlap: 15 own a watch and 7 also a headset means watch-only = $15 - 7 = 8$ .

A probability is region ÷ universal set. The denominator is the whole group surveyed. With 60 people and 7 owning both, $P(\text{both}) = 7/60$ — never $7/39$ over a region.

A conditional fixes the subgroup as the denominator. “A random child” from 80 children, 30 of whom chose pizza, gives $30/80$ , not $30/120$ : the universal set is whatever group you are choosing from.

Worked example

37 people are surveyed: 20 can be a trainer (3 of them do both jobs), 15 can do reception (the same 3), and 5 can do neither. Work out the probability a random person can be a trainer.

Split each set into regions. Trainer-only = $20 - 3 = 17$ , reception-only = $15 - 3 = 12$ , both = 3, neither = 5.
$17 + 12 + 3 + 5 = 37$
Use the WHOLE trainer set on top. The trainer set is the trainer-only region plus the overlap: $17 + 3 = 20$ .
Divide by the universal set.
$P(\text{trainer}) = \frac{20}{37}$

The traps: $17/37$ drops the 3 in the overlap from the trainer set, $20/32$ drops the 5 who can do neither from the whole, and $17/20$ uses the trainer set as the whole group. A set total includes its overlap, and a probability is over the universal set.

Find out if this is costing you marks

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Common questions

Why isn’t the watch-only group just the number written by the watch?: Because that number is the whole watch set — it already counts the people who also own a headset. To get watch-only you take the overlap back out: $15 - 7 = 8$ , and $8 + 7 = 15$ confirms it.
What goes on the bottom of a probability from a Venn diagram?: The universal set — the total number of people surveyed. With 60 people and 7 owning both items, $P(\text{both}) = 7/60$ . A region count like 39 is the wrong denominator.
A conditional asks “of the children” — do I still divide by the total?: No. The denominator is the group the question names. If 30 of 80 children chose pizza, then the probability a random child chose pizza is $30/80$ , not $30/120$ .
How do I check a Venn diagram is read correctly?: Add up every region — the “only” parts, the overlap, and the outsiders — and confirm they sum to the universal set: $17 + 12 + 3 + 5 = 37$ . If they don’t, a region has been miscounted.

Related misconceptions

Combined-event probability: why you multiply along a branch, not add the probabilitiesThe neighbouring probability misconception — the same 'which numbers combine, and how' confusion, met on a tree diagram instead of a Venn.
Relative frequency and expected outcomes: why 'expected' is P × trials, not a tidy round numberThe third probability vertical — once you can read a count from a diagram, you turn it into an estimate of probability and an expected frequency.

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