Venn diagrams: sets, notation and probability

$\xi = \{1, 2, 3, \ldots, 12\}$, A = multiples of 3, B = even numbers.

• $A = \{3, 6, 9, 12\}$ and $B = \{2, 4, 6, 8, 10, 12\}$.
• $A \cap B = \{6, 12\}$: these go in the overlap.
• A only: 3 and 9. B only: 2, 4, 8 and 10.
• Outside both circles: 1, 5, 7 and 11.

So $A \cup B = \{2, 3, 4, 6, 8, 9, 10, 12\}$, and $A' = \{1, 2, 4, 5, 7, 8, 10, 11\}$.

In a class of 30 students, 18 play football, 12 play tennis and 5 play both.

1. Overlap first: 5 play both.
2. Football only: $18 - 5 = 13$.
3. Tennis only: $12 - 5 = 7$.
4. Outside: $30 - (13 + 5 + 7) = 5$ play neither.

A common mistake is putting 18 and 12 straight into the circles, which counts the 5 students twice.

Divide the number in the region you want by the total. Using Example 2:

• $P(\text{plays tennis but not football}) = \tfrac{7}{30}$.
• $P(F \cup T) = \tfrac{13 + 5 + 7}{30} = \tfrac{25}{30} = \tfrac{5}{6}$.
• $P(\text{neither}) = \tfrac{5}{30} = \tfrac{1}{6}$.

Conditional probability (higher tier): "given that a student plays tennis, find the probability they also play football". Only look at the tennis circle: $\tfrac{5}{12}$. The word "given" changes the denominator.

Some questions give expressions instead of numbers. 50 people were asked. A only contains $x$, the overlap is 8, B only is $2x$ and 6 are outside both circles.

So 12 are in A only and 24 in B only. Always check the regions add up to the total.

With three sets, there's a centre region (in all three), three regions in exactly two sets, three regions in one set only, and the outside. Work from the centre outwards: fill the middle, then the pairwise overlaps (subtracting the centre), then the single regions.

• Forgetting the region outside the circles.
• Mixing up $\cap$ (and) and $\cup$ (or). A memory trick: $\cup$ looks like a cup that holds everything in either set.
• Using the whole total as the denominator in a "given that" question.

1. 40 people: 22 have a cat, 15 have a dog, 6 have both. How many have neither?
2. Using question 1, find P(dog only).
3. $\xi = \{1, \ldots, 10\}$, A = odd numbers, B = prime numbers. List $A \cap B$.

Answers: 1. Cat only 16, dog only 9, both 6, so $40 - 31 = 9$ have neither. 2. $\tfrac{9}{40}$. 3. $\{3, 5, 7\}$ (2 is prime but even; 1 is not prime).

More practice in our Venn diagram questions.