Inclusion-exclusion principle for two sets

The inclusion-exclusion principle clarifies the problem of finding the number of elements in the union of two sets that may have elements in common.

We can use the notation $#A$#A and $#B$#B to mean the counts of elements in sets $A$A and $B$B respectively. If we want to know the value of $#(A\cup B)$#(A∪B) we might get an initial estimate by adding $#A+#B$#A+#B. 

But, any elements that are in both $A$A and $B$B would have been counted twice. Therefore we conclude that the correct result should be 

$#(A\cup B)=#A+#B-#(A\cap B)$#(A∪B)=#A+#B−#(A∩B)

This formula occurs in probability calculations.

In another chapter, we show how it extends to situations involving more than two sets.

 

Example 1

In the Venn diagram above, set $A$A has $13$13 elements and set $B$B has $10$10 elements. The three red dots are the elements that are common to both $A$A and $B$B.

You could check by counting that $#(A\cup B)=#A+#B-#(A\cap B)$#(A∪B)=#A+#B−#(A∩B). That is, $#(A\cup B)=13+10-3=20$#(A∪B)=13+10−3=20.

 

If sets $A$A and $B$B are interpreted to be events and the dots in the diagram are the individual outcomes of a sequence of repetitions of an experiment, then we can make statements about the proportions of outcomes belonging to each event and hence, we can make inferences about their probabilities.

In this experiment, there were $31$31 trials. Hence, event $A$A occurred in the proportion $\frac{13}{31}$1331​, event $B$B occurred in the proportion $\frac{10}{31}$1031​. Events $A$A and $B$B occurred together in $\frac{3}{31}$331​ of the trials. We can deduce that the proportion of trials in which event $A$A or event $B$B occurred was $\frac{13+10-3}{31}=\frac{20}{31}$13+10−331​=2031​.

 

Example 2

In a survey of an animal population, it was found that individuals could have either light or dark colouring and they could have wide or narrow skull shape. The light coloured animals represented 35% of the surveyed group. The individuals with wide skulls represented 55% of the group. Animals with either wide skulls or light colouring made up 81% of the group.

What proportion had both light colouring and a wide skull?

Let $A$A be the set of light coloured animals and let $B$B be the set of animals with wide skulls. If the number of animals surveyed was $t$t, then the proportion with one or other of the features can be written $\frac{#(A\cup B)}{t}=\frac{#A+#B-#(A\cap B)}{t}$#(A∪B)t​=#A+#B−#(A∩B)t​. Then,

$\frac{#A+#B-#(A\cap B)}{t}$#A+#B−#(A∩B)t​	$=$=	$\frac{#A}{t}+\frac{#B}{t}-\frac{#(A\cap B)}{t}$#At​+#Bt​−#(A∩B)t​
$0.81$0.81	$=$=	$0.35+0.55-\frac{#(A\cap B)}{t}$0.35+0.55−#(A∩B)t​
$\frac{#(A\cap B)}{t}$#(A∩B)t​	$=$=	$0.35+0.55-0.81$0.35+0.55−0.81
	$=$=	$0.09$0.09

Thus, we conclude that $9%$9% of the animals had both features.