Naming the Regions of a Venn Diagram

Four regions

A survey asks 18 homes whether they keep a dog and whether they keep a cat. A Venn diagram sorts them. The box stands for all 18 homes. One circle holds the homes with a dog and the other holds the homes with a cat, and the circles overlap.

The two circles cut the box into four regions, and every home sits in exactly one of them: a dog and no cat, a dog and a cat, a cat and no dog, or neither pet. Here those regions hold 6, 3, 5 and 4 homes, and 6 + 3 + 5 + 4 = 18, every home counted once.

The 18 homes in four regions: 6 with only a dog, 3 with both pets, 5 with only a cat and 4 with neither.

The overlap

The overlap is the part inside both circles at once. It holds the 3 homes that keep a dog and a cat. A home with both pets is not drawn twice, once in each circle; it is drawn once, in the place where the circles cross.

The overlap holds the 3 homes that keep a dog and a cat.

One circle alone

The dog circle without the overlap holds the 6 homes that keep a dog and no cat. This region is called "dog only".

The whole dog circle is "dog only" together with the overlap, so 6 + 3 = 9 homes keep a dog. In the same way 5 homes keep only a cat, and 5 + 3 = 8 homes keep a cat. "Keeps a dog" means the whole circle, 9 homes; "keeps only a dog" means 6.

The dog circle outside the overlap: 6 homes with a dog and no cat.

Outside both circles

The part of the box outside both circles holds the 4 homes that keep neither pet. They are still part of the survey, which is why the box is drawn: it is the universal set, everything being sorted.

Everything outside one circle is called the complement of that set. The complement of "dog", written dog′, is the cat-only homes and the homes with neither: 5 + 4 = 9 homes without a dog. And 9 with a dog plus 9 without is all 18.

Outside both circles but inside the box: 4 homes with neither pet.

Intersection and union

Two symbols name the regions made by two sets A and B. $A\cap B$A ∩ B, the intersection, is the overlap: everything in A and in B. $A\cup B$A ∪ B, the union, is everything inside either circle: in A or in B, or in both.

With A for dog and B for cat, $A\cap B$A ∩ B holds 3 homes. $A\cup B$A ∪ B holds the three regions inside the circles, 6 + 3 + 5 = 14 homes. Adding the circle totals instead, 9 + 8 = 17, counts the 3 homes in the overlap twice, so take them off once: 17 − 3 = 14.

The union, dog $\cup$∪ cat: every region inside a circle, 6 + 3 + 5 = 14 homes.

Filling a diagram from totals

Questions usually give the circle totals, not the regions. In a class of 30, 12 students play soccer, 10 play tennis and 4 play both. Fill the overlap first: 4. Then each circle alone: 12 − 4 = 8 play only soccer and 10 − 4 = 6 play only tennis. The circles hold 8 + 4 + 6 = 18 students, so 30 − 18 = 12 play neither.

Check by adding the four regions back to the class: 8 + 4 + 6 + 12 = 30. Adding the two totals, 12 + 10 = 22, overstates the 18 who play a sport, because the 4 who play both are in both totals.

Two circles of 12 and 10, here with 4 in the overlap. Slide the overlap: each "only" region loses what the overlap gains, and the union is always 12 + 10 minus the overlap.

Probabilities from regions

Pick one of the 18 homes at random, so each home is equally likely. Each probability is a region's count over 18. The chance the home keeps both pets is $\frac{3}{18}=\frac{1}{6}$3/18 = 1/6. The chance it keeps a dog is $\frac{9}{18}=\frac{1}{2}$9/18 = 1/2. The chance it keeps a dog or a cat is $\frac{14}{18}=\frac{7}{9}$14/18 = 7/9, and the chance it keeps neither is $\frac{4}{18}=\frac{2}{9}$4/18 = 2/9. The last two add to 1, because neither is the complement of the union.

If the home is picked from the dog owners only, the whole is the 9 homes in the dog circle, not 18. Then the chance it also keeps a cat is $\frac{3}{9}=\frac{1}{3}$3/9 = 1/3.

The usual mistakes

Writing a circle's total in its "only" region. If 12 play soccer and 4 play both, the soccer-only region holds 8; writing 12 there counts the 4 twice.

Adding the circle totals to find $A\cup B$A ∪ B. The overlap is inside both circles, so it must be counted once.

Leaving out the region outside the circles. The four regions together must add to the whole, including the ones in neither set.

Reading "has a dog" as "has only a dog". The homes with both pets have a dog too.

Music and drama

In the application below, the totals for music and drama in a year group of 60 are written into a Venn diagram, the overlap first, and two probabilities are read from it: one out of the whole year group, and one out of the music students only.