Probability in Set Notation

Twenty outcomes in four regions

An experiment has 20 outcomes, all equally likely. An event is a set of some of those outcomes, so two events A and B can be drawn as two circles in a Venn diagram, with the box around them standing for all 20.

Here 6 outcomes are in A only, 4 are in both A and B, 5 are in B only, and 5 are in neither. Check: 6 + 4 + 5 + 5 = 20, every outcome counted once.

The probability of an event is the number of outcomes in it divided by 20. A holds 6 + 4 = 10 outcomes, so $P\left(A\right)=\frac{10}{20}=\frac{1}{2}$P(A) = 10/20 = 1/2. B holds 4 + 5 = 9, so $P\left(B\right)=\frac{9}{20}$P(B) = 9/20.

The 20 outcomes: 6 in A only, 4 in both, 5 in B only and 5 in neither.

And: the intersection

$A\cap B$A ∩ B, the intersection, is the set of outcomes in A and in B at the same time. On the diagram it is the overlap, which holds 4 outcomes.

So $P\left(A\cap B\right)=\frac{4}{20}=\frac{1}{5}$P(A ∩ B) = 4/20 = 1/5. The 6 outcomes in A only and the 5 in B only are left out, because each of them is in one event and not the other.

$A\cap B$A ∩ B is the overlap: 4 of the 20 outcomes, so $P\left(A\cap B\right)=\frac{4}{20}$P(A ∩ B) = 4/20.

Or: the union

$A\cup B$A ∪ B, the union, is the set of outcomes in A or in B or in both. In everyday English "tea or coffee" can mean one and not the other; in set notation "or" always includes both. On the diagram the union is all three regions inside the circles: 6 + 4 + 5 = 15 outcomes.

So $P\left(A\cup B\right)=\frac{15}{20}=\frac{3}{4}$P(A ∪ B) = 15/20 = 3/4.

Adding P(A) and P(B) gives $\frac{10}{20}+\frac{9}{20}=\frac{19}{20}$10/20 + 9/20 = 19/20, which is too much: the 4 outcomes in the overlap are in A and in B, so they were added twice. Subtract them once: $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=\frac{10}{20}+\frac{9}{20}-\frac{4}{20}=\frac{15}{20}$P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 10/20 + 9/20 − 4/20 = 15/20, the same answer as the count.

$A\cup B$A ∪ B is every region inside a circle: 6 + 4 + 5 = 15 of the 20 outcomes.

Not: the complement

A', the complement of A, is every outcome not in A: the 5 in B only and the 5 in neither, 10 in all. So $P\left(A^{\prime }\right)=\frac{10}{20}$P(A') = 10/20. An outcome is either in A or not in A, so $P\left(A^{\prime }\right)=1-P\left(A\right)=1-\frac{10}{20}=\frac{10}{20}$P(A') = 1 − P(A) = 1 − 10/20 = 10/20.

${\left(A\cup B\right)}^{\prime }$(A ∪ B)' is every outcome outside both circles: the 5 in neither. So $P\left({\left(A\cup B\right)}^{\prime }\right)=\frac{5}{20}=\frac{1}{4}$P((A ∪ B)') = 5/20 = 1/4, which is $1-\frac{15}{20}$1 − 15/20.

${\left(A\cup B\right)}^{\prime }$(A ∪ B)' is the part of the box outside both circles: 5 of the 20 outcomes.

A only

"In A but not in B" is $A\cap B^{\prime }$A ∩ B': in A, and in the complement of B. On the diagram it is the part of circle A outside the overlap, which holds 6 outcomes, so $P\left(A\cap B^{\prime }\right)=\frac{6}{20}=\frac{3}{10}$P(A ∩ B') = 6/20 = 3/10.

It is also A with the overlap taken away: $P\left(A\right)-P\left(A\cap B\right)=\frac{10}{20}-\frac{4}{20}=\frac{6}{20}$P(A) − P(A ∩ B) = 10/20 − 4/20 = 6/20.

$A\cap B^{\prime }$A ∩ B' is circle A outside the overlap: 6 of the 20 outcomes.

From a list to a diagram

Pick one of the whole numbers 1 to 10 at random. Let A be the even numbers, {2, 4, 6, 8, 10}, and B the numbers up to 4, {1, 2, 3, 4}.

$A\cap B=$A ∩ B = {2, 4}, so $P\left(A\cap B\right)=\frac{2}{10}=\frac{1}{5}$P(A ∩ B) = 2/10 = 1/5. $A\cup B=$A ∪ B = {1, 2, 3, 4, 6, 8, 10}, seven numbers, so $P\left(A\cup B\right)=\frac{7}{10}$P(A ∪ B) = 7/10. A' = {1, 3, 5, 7, 9}, so $P\left(A^{\prime }\right)=\frac{5}{10}=\frac{1}{2}$P(A') = 5/10 = 1/2.

Check the union with the rule: $P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=\frac{5}{10}+\frac{4}{10}-\frac{2}{10}=\frac{7}{10}$P(A) + P(B) − P(A ∩ B) = 5/10 + 4/10 − 2/10 = 7/10. In the diagram, 6, 8 and 10 are in A only, 2 and 4 are in both, 1 and 3 are in B only, and 5, 7 and 9 are in neither: 3 + 2 + 2 + 3 = 10.

The numbers 1 to 10: 3 even numbers above 4, the 2 even numbers up to 4, 2 odd numbers up to 4, and 3 numbers in neither set. The union holds 3 + 2 + 2 = 7 of them.

The usual mistakes

Counting only the overlap for $P\left(A\cup B\right)$P(A ∪ B). The union takes the outcomes in one set only as well: $\frac{15}{20}$15/20, not $\frac{4}{20}$4/20.

Counting both circles for $P\left(A\cap B\right)$P(A ∩ B). The intersection wants only the outcomes in both sets at once: $\frac{4}{20}$4/20, not $\frac{15}{20}$15/20.

Adding P(A) and P(B) without taking off the overlap. $\frac{10}{20}+\frac{9}{20}=\frac{19}{20}$10/20 + 9/20 = 19/20 counts the 4 shared outcomes twice.

Adding the outcomes outside the circles to the total again. The 20 already includes the 5 in neither, so the denominator stays 20.

Two mailing lists

In the application below, a shop's customers are sorted by two lists, A and B. The customers who receive an email are $A\cup B$A ∪ B, and those who receive none are ${\left(A\cup B\right)}^{\prime }$(A ∪ B)'. Chosen at random from all 2500 customers, a customer receives the email with probability $\frac{1790}{2500}$1790/2500 and is on both lists with probability $\frac{310}{2500}$310/2500.