A union is the area that belongs to one or both of two events. An intersection is the area that belongs to both of those two events. Below is the notation:
The union of two events is written as follows: \(A \cup B\) The intersection of two events is written as follows: \(A \cap B\)
Based on the definitions, the union of two events ALWAYS includes its intersection.
Here is one other piece of notation that will be used that you should be familiar with.
The notation $$[]$$ refers to the "size" of something. It is usually used to indicate the size of an event, as in how many different outcomes can lead to the event.
These two different concepts are used in different calculations, within various applications in probability and statistics.
From here, a number of examples will be provided to give a better understanding of union calculations.
Example 1: Of the customers who bought items at a store on a particular day, \(53\) of them bought clocks and \(47\) of them bought watering cans. What is the largest possible number of people that bought either a clock or a watering can on that day?
Solution: Let \(W\) be the event that a person bought a watering can, and let \(C\) be the event that a person bought a clock. Then:
$$[W \cup C] = [W] + [C] - [W \cap C]$$
To maximize the size of the union, the size of the intersection must be minimized. There are no further restrictions on the intersection found in the context of the problem, so just let it be \(0\), giving
$$[W \cup C] = 53 + 47 = 100$$
So up to 100 people could have bought clocks or watering cans at the store that day.
The next two examples lack context in order to emphasize the algebraic manipulations.
Example 2: If \([A \cap B] = 75\), \([A] = 143\), and \([B] = 173\), find \([A \cup B]\).
Solution: Use the formula for the union of two events:
$$[A \cup B] = [A] + [B] - [A \cap B] = 143 + 173 - 75 = 241$$
Example 3: If \([A] = 45\), \([B] = 34\), and \([A \cup B] = 51\), find the value of \([A \cap B]\).
Solution: The equation relating unions and intersections will again be used, but in a slightly different manner than in the previous example.
$$[A \cup B] = [A] + [B] - [A \cap B] \Rightarrow$$ $$51 = 45 + 34 - [A \cap B] \Rightarrow$$ $$[A \cap B] = 79 - 51 = 28$$
Notice here, the equation had to be solved for the desired value.
The process used to solve for the variable in the previous example can be done in a general form.
Example 4: Use the equation for the union to write an expression in terms of the intersection and of \([A] + [B]\).
Solution: We will start with the equation for the union:
$$[A \cup B] = [A] + [B] - [A \cap B]$$
Now solve this equation for the intersection:
$$[A \cup B] + [A \cap B] = [A] + [B]$$
Here is the expression for the sum of the two standard events' sizes. Now isolate \([A \cap B]\):
$$[A \cap B] = [A] + [B] - [A \cup B]$$
These expressions are easy to re-derive anytime you need them.
Example 5: The number of people in an auditorium with a graduate degree and at least 2 children is 94. The number of people in an auditorium with a graduate degree or have at least two children is 331. If \(G\) and \(C\) are these two events respectively, find \([G] + [C]\).
Solution: From one of the equations derived in Example 3:
$$[G] + [C] = [G \cap C] + [G \cup C] = 94 + 331 = 425$$
Unions and intersections also have an application to probability, which is part of the AP Statistics curriculum. The concepts and calculations for unions and intersections can be performed perfectly validly with probabilities. The next example illustrates this.
Example 6: If \(P(A) = 0.4\) and \(P(B) = 0.35\), write an expression for \(P(A \cap B)\) in terms of \(P(A \cup B)\). What is the largest possible value that \(P(A \cup B)\) can have?
Solution: The formulas presented earlier in the article are acceptable to apply to probabilities:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Plug in the known information:
$$P(A \cup B) = 0.4 + 0.35 - P(A \cap B) \Rightarrow$$ $$P(A \cup B) = 0.75 - P(A \cap B) \Rightarrow$$ $$P(A \cup B) + P(A \cap B) = 0.75 \Rightarrow$$ $$P(A \cap B) = 0.75 - P(A \cup B)$$
An intersection of two events can not contain more of anything than either event does by itself, so the maximum size of the intersection is the size of the smallest event in the intersection. Therefore, we use the event that is least likely to happen (putting things in the perspective of probabilities), which is event \(B\). Therefore the maximum value of \(P(A \cap B)\) is \(0.35\).
Here is one additional example.
Example 7: A bus goes through a certain route every day to take people to and from work. The probability of a rock hitting the windshield on any particular day is \(0.04\). The probability of a flat tire occurring is \(0.05\). The probability of both events happening is \(0.003\). What is the probability that at least one of the events will happen on a particular day?
Solution: Let \(R\) be the event of the windshield getting hit with a rock. Let \(F\) be the probability of getting a flat tire. Then use the equation involving the union and intersection of two events:
$$P(R \cup F) = P(R) + P(F) - P(R \cap F) = 0.04 + 0.05 - 0.003 = 0.087$$
Mutually exclusive events are a common sight in probability-related scenarios. Such events cannot both occur simultaneously.
If two events are mutually exclusive, then the intersection of these events has a size or probability of 0.
If these events are \(A\) and \(B\), then
$$[A \cap B] = 0$$ $$P(A \cap B) = 0$$
Which identity you use depends on the context of the problem.
The next example derives an important formula for mutually exclusive events.
Example 8: If the intersection of two events \(A\) and \(B\) is \(0\), then find the union of these events in terms of the two events.
Solution: We have
$$[A \cup B] = [A] + [B] - [A \cap B]$$
The intersection is \(0\), so substitute that in:
$$[A \cup B] = [A] + [B] - 0 = [A] + [B]$$
The above example could have been done exactly the same with probabilities, and both forms work exactly the same way, except that once again context determines which formula should be used.
Example 9: A very unusual die with 7 faces is rolled one; each face has an equal probability of being landed on. Two of the sides have a “3,” two of the sides have a “4,” and the other sides are blank. Let these three respective events be represented by \(T\), \(F\), and \(B\). Find \(T \cup F\) and \(T \cap F\).
Solution: The die cannot land on two faces at the same time, so all three events are mutually exclusive. Therefore, by definition, \(P(T \cap F) = 0\), and the union is calculated as follows:
$$P(T \cup F) = P(T) + P(F) = \frac{2}{7} + \frac{2}{7} = \frac{4}{7}$$
Obviously the concept of mutually exclusive events has different applications in real life, and as you continue studying probability and/or statistics, you will realize how common unions and intersections truly are, so master the basic concepts now.