A Venn diagram is used to sort information about which elements belong to which sets. Anything that can be organized in a Venn diagram can also be organized in a contingency table. Most of the time you can choose which of them to use. Making a Venn diagram can sound like a lot of work, but it’s way easier than it may seem.

Theory

A Venn diagram is a graphic representation of multiple sets. The diagram gives a good overview of how many elements are in each set, how many elements the different sets have in common, and the total number of elements. A Venn diagram will also show you if there are elements that aren’t part of any sets. Unions, intersections and disjoint sets are very clearly presented in a Venn diagram.

Example 1

**A group of girls are preparing for prom, and decide to go dress shopping together. Out of the 32 girls, 20 of them buy short dresses, while 25 buy long dresses. What is the probability of a girl chosen at random having bought both a short and a long dress? **

In assignments like these, it pays off to draw a Venn diagram. In that case, you have to know how many girls are in the intersection. You find this by adding together all the dresses you know the girls have bought and then subtracting the total number of girls.

$$\text{Intersection/both}=20+25-32=13$$ |

From this, you can now find out how many only bought a short dress and how many only bought a long dress by subtracting the intersection from each group. This gives you:

$$\begin{array}{llll}\hfill \text{Onlyshortdress}& =20-13=7\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{Onlylongdress}& =25-13=12\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Now you can put this information in a Venn diagram. It looks like this:

When you now want to find the probability of a girl picked at random having bought both kinds of dresses, you use the formula for uniform probability, because you assume the probability of a girl buying a short dress is the same as the probability of her buying a long dress. Then the calculation becomes

$$\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(\text{buysboth}\right)& =\frac{13}{32}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.406\phantom{\rule{0.17em}{0ex}}25\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 40.63\phantom{\rule{0.17em}{0ex}}\text{\%}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$which means the probability of a girl picked at random bought both a long dress and a short dress is $40.63$ %.