Union means joining or combination of two or more things. The union of two sets A and B is the set of elements in A or B (or both). 

Example: The union of the "Soccer" and "Tennis" sets is janice, hunter, christy, nathan and josh

Soccer ∪ Tennis = {janice, hunter, christy, nathan and josh}

An event is the fact that something has happened as an outcome of an experiment like throwing a ball, tossing a coin etc. Likewise, getting a four when a dice is thrown, getting a head etc are some events.

The union is one of the most-commonly performed operations on events. As the name suggests, the word "union" means to join two things irrespective of how they behave individually. The union of events means occurrence of either of the given events. This implies it is a must for only one of them is to occur which can be anyone of the given events, but one of them is a must to occur.
There are two ways to represent the union of two or more sets. We make use of either ‘$\cup$’ a “u -shaped symbol” representing union of two or more sets.

Another way is Logical “OR”. The OR written in between the events to be taken union of also implies the same as the U-shaped symbol.
Let A and B be two events. Then in general, A is representing the set of all outcomes that are favorable to the occurrence of the event A and similarly B represents the set of all outcomes that are favorable to B. Now A $\cup$ B is the set of all outcomes that are either in A or in B, that is A $\cup$ B is the super set of A and B

Clearly A $\subseteq$ A $\cup$ B and B $\subseteq$ A $\cup$ B.

Every event is associated to an experiment.
When we take union of two or more events it depends on the type of events as to which formula to use. To find the probability of an event we use the formula below:

P (E) = $\frac{number\ of\ favorable\ outcomes}{total\ number\ of\ outcomes}$
Let A and B be two events, P (A) and P (B) be the individual probabilities of the events A and B respectively. Then, P (A $\cup$ B) is the probability of the union of the given two sets. We can have two types of cases:

1) The events are mutually exclusive:

In this case the given events have no common outcome. For example consider the experiment of throwing a dice. Getting a prime number and getting a 4 are two mutually exclusive events. In both cases there is no outcome that is common. So these two are mutually exclusive events. 

In this case we use the formula below for finding the probability of the events whenever two out of three values are known
P (A $\cup$ B) = P (A) + P(B)

2) The events are not mutually exclusive:

In this case the given events have some common outcomes. To avoid the duplicacy of outcomes in calculation of final probability we then follow the formula below:
P (A $\cup$ B) = P (A) + P (B) - P (A $\cap$ B)

Here, A $\cap$ B is the set of common outcomes of the events A and B and P (A $\cap$ B) is the probability of the getting these common outcomes.

The above formulas can be used to find any quantity out of the three or four if given the remaining ones. Also, it is important to first see and judge what types of events are given to us and then accordingly move forward with the compound set of union. We call it compound as we are joining here two sets with the condition of either.

Also, the union of two complement events is always equal to 1. That is, P (A $\cup$ A’) = 1

Where A’ is the complement event of A. We can prove this by considering  the event of getting an even number when we throw a dice. This implies P (A) = $\frac{1}{2}$. Now, the complement of event A here is getting an odd number. This implies P (A’) = $\frac{1}{2}$. Also we know that the two complement sets are always mutually exclusive of each other. This implies, P (A $\cup$ A’) = P (A) + P (A’) = $\frac{1}{2}$ + $\frac{1}{2}$ = 1. 

We can take any example to show this and the result will always be the same.
Let us see some examples to understand the concept of union of events in a better manner.

Example 1:

Find the probability of given:

a) Getting a four or an even number when a dice is thrown

b) Picking a king or an ace from the deck of 52 cards.

Solution:

a) Let A = getting a four. $\Rightarrow$ P (A) = $\frac{1}{6}$

Let B = even number $\Rightarrow$ P (B) = $\frac{3}{6}$

A $\cap$ B = {4} $\Rightarrow$ P (A $\cap$ B) = $\frac{1}{6}$

Now, P (A $\cup$ B) = P (A) + P (B) - P (A $\cap$ B)

= $\frac{1}{6}$ + $\frac{3}{6}$ - $\frac{1}{6}$ = $\frac{3}{6}$ = $\frac{1}{2}$

b) Let A = king $\Rightarrow$ P (A) = $\frac{4}{52}$ = $\frac{1}{13}$

Let B = ace $\Rightarrow$ P (B) = $\frac{4}{52}$ = $\frac{1}{13}$

A $\cap$ B = NULL

Now, P (A $\cup$ B) = P (A) + P (B) = $\frac{2}{13}$