# Probability with Venn Diagrams – Ultimate Guide

Visualising multiple events using Venn diagrams to find probabilities is done quite often.

## Representing Data in Venn Diagrams

There are 9 types of Venn diagrams for holding two events when they overlap each other. Let’s assume there are two elective subjects, Arts and Biology, in a certain school.

$P(A)$
The probability that a randomly chosen student selected Arts

$P(B)$
The probability that a randomly chosen student selected Biology

$P(only \ A)$
The probability that a randomly chosen student selected only Arts

$P(only \ B)$
The probability that a randomly chosen student selected only Biology

$P(not \ A)$
The probability that a randomly chosen student did not select Arts

$P(not \ B)$
The probability that a randomly chosen student did not select Biology

$P(not \ A \ nor \ B)$
The probability that a randomly chosen student did not select both

$P(A \ or \ B)$
The probability that a randomly chosen student selected Arts or Biology

$P(A \ and \ B)$
The probability that a randomly chosen student selected Arts and Biology

Consider one of the probability rules: $P(A \ or \ B) = P(A) + P(B) – P(A \ and \ B)$. A Venn diagram also uses this probability formula to find the demanded probability.

## Example 1

Find the probability given the Venn diagram is given.

(a) $P(A) \displaystyle = \frac{a \ b \ c}{a \ b \ c \ d \ e} = \frac{3}{5}$

(b) $P(none) = \displaystyle \frac{e}{a \ b \ c \ d \ e} = \frac{1}{5}$

(c) $P(A \ not \ B) = \displaystyle \frac{c}{a \ b \ c \ d \ e} = \frac{1}{5}$

(d) $P(A \ and \ B) = \displaystyle \frac{a \ b}{a \ b \ c \ d \ e} = \frac{2}{5}$

(e) $P(A \ or \ B) = \displaystyle \frac{a \ b \ c \ d}{a \ b \ c \ d \ e} = \frac{4}{5}$

Now you can watch the video lesson below for further study.

## Representing Events without Replacements in Venn Diagrams

Probability without replacement indicates once the first item was drawn, then the item is not to be put back into the sample space before drawing a second item.

## Example 2

Two elements are chosen at random without replacement. Find the probability given the Venn diagram is given.

(a) $P(AA) = \displaystyle \frac{4}{7} \times \frac{3}{6} = \frac{2}{7}$

(b) $P(AB) = \displaystyle \frac{4}{7} \times \frac{3}{6} = \frac{2}{7}$

(c) $P(BB) = \displaystyle \frac{3}{7} \times \frac{2}{6} = \frac{1}{7}$

(d) $P(not \ B \ and \ B) = \displaystyle \frac{4}{7} \times \frac{3}{6} = \frac{2}{7}$

## Probabilities in Venn Diagrams

The outcomes of an event can be represented using a Venn diagram. Rather the sample space is represented in the Venn diagram, and often, the probabilities are given in the Venn diagram.

## Drawing Three Circles in Venn Diagrams

Three-circle Venn diagrams are an advanced step in complexity from two-circle Venn diagrams. The following Venn diagrams illustrate the distinct differences between the regions selected.

$P(C)$

$P(not \ C)$

$P(none)$

$P(A \ not \ C)$

$P(C \ not \ B)$

$P(B \ not \ A)$

$P(A \ and \ C)$

$P(A \ and \ B)$

$P(B \ and \ C)$

$P(A \ and \ B \ not \ C)$

$P(A \ and \ C \ not \ B)$

$P(B \ and \ C \ not \ A)$

$P(A \ and \ B \ and \ C)$

$P(C \ not \ A \ not \ B)$

## Example 3

Find the probability given the Venn diagram where the element numbers are given.

(a) $P(A) = \displaystyle \frac{1+4+5+7}{1+2+3+4+5+6+7+8} = \frac{17}{36}$

(b) $P(not \ B) = \displaystyle \frac{1+5+3+8}{1+2+3+4+5+6+7+8} = \frac{17}{36}$

(c) $P(A \ and \ B \ and \ C) = \displaystyle \frac{7}{1+2+3+4+5+6+7+8} = \frac{7}{36}$

(d) $P(A \ not \ B) = \displaystyle \frac{1+5}{1+2+3+4+5+6+7+8} = \frac{6}{36} = \frac{1}{6}$

## Conditional Probabilities with Venn Diagrams

To calculate the conditional probability of event A given event B, identify the region representing $P(A \ and \ B)$ and $P(B)$, and this is denoted as

$\displaystyle P(A|B) = \frac{P(A \ and \ B)}{P(B)}$.

## Example 4

Find the following probabilities.

(a) $P(A|B)$

$P(A|B) = \displaystyle \frac{P(A \ and \ B)}{P(B)} = \frac{\{d\}}{\{b \ d \ f\}} = \frac{1}{3}$

(b) $P(B|A)$

$P(B|A) = \displaystyle \frac{P(A \ and \ B)}{P(A)} = \frac{\{d\}}{\{a \ d \ e\ g\}} = \frac{1}{4}$