Union of Events Examples - Word Problems
The word ‘union’ means combining of two or more groups and expressing the sum of the total items under one heading. We do not go into the details of the numbers in each group. For example let one basket $A$ contains apples and oranges and another basket $B$ contains pineapples and bananas. Suppose we put all the contents of both the baskets in a third basket $C$, we say that the basket $C$ contains all the four fruits apples, oranges, pineapples and bananas. Then we can call that basket $C$ is a union of baskets $A$ and $B$.  Mathematically it is denoted as $C$ = $A \cup B$. We did not talk about the number of any of the fruits. In fact in certain cases a common item may exist in both the groups but when combined it becomes only one item. For example the union of one basket containing apples and oranges with another basket of oranges and bananas is only described as a group of apples, oranges and bananas and not as apples, oranges, oranges and bananas. This is the concept used in set theory which covers not only to physical items but also the events. Its applications are helpful in probability problems.

Union of Events

With the basic knowledge of what a union is, we will do a closer study on union of events. An event is something could happen in a random experiment means it is one of the options in the sample space. In an experiment of tossing a coin getting a head or getting a tail are events but getting a $2$ is not because it is not in the sample space. Therefore, a union of events in a random experiment is the condition that at least one of the combined events occurs. To understand clearer let us study an example. Suppose we roll a fair die and call event $A$ and event $B$ as getting a prime number and a multiple of $3$ respectively. Then, the set $A$ is $\{2,3,5,\}$ and the set $B$ is $\{3, 6\}$ Then $A \cup B$ is represented by the set $\{2,3,5,6\}$. If you roll on the numbers $2, 3$ and $5$ we can say event $A$ has occurred and if you roll on $6$ we can say event $B$ has occurred. If roll on $3$, both the events $A$ and $B$ have occurred jointly. The element $3$ which occurs in both the sets $A$ and $B$ is called the intersection of set $A$ and set $B$ and denoted as $A \cap B$. 

In general if '$n$' denotes the respective number of events, then,

$n\ (A \cup B)$ = $n(A)\ +\ n(B)\ - n(A \cap B)$

This analysis is very helpful in probability of union of two events. The formula used in probability is, 

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

In special cases, occurrence of both events $A$ and $B$ is not possible. For example in tossing a coin you cannot have both 'tossing on head' and 'tossing on tail'. Such events are called as mutually exclusive events. In such case $P \cap B$ = $0$ and the formula reduces to,

$P\ (A \cup B)$ = $P(A) + P(B)$

The formula can be extended for any number of events with a little modification. For example, in case of three sets $A, B$ and $C$,
$n\ (A \cup B \cup C)$ = $n\ (A)\ +\ n\ (B)\ +\ n\ (C)\ -\ n\ (A \cap B)\ -\ n (B \cap C)\ -\ n\ (C \cap A)\ +\ n\ (A \cap B \cap C)$

Word Problems

Example 1: 

Consider the same problem as exampled earlier. If you roll a fair die, what is the probability of rolling on a prime number or on a multiple of $3$. 


When a fair die is rolled, all possible outcomes are all the numbers from $1$ to $6$. That is, the sample space set $S$ is $\{1, 2, 3, 4, 5, 6\}$ and hence $n(S)$ = $6$

Let us call rolling on a prime number as event $A$ and rolling on multiple of $3$ as event $B$. The set for event $A$ is $\{2, 3, 5\}$, the numbers which are the only prime numbers in the sample space and the set for event $B$ is $\{3, 6\}$, the only numbers that are multiples of $3$ in the sample space. So, $n(A)$ = $3$ and $n(B)$ = $2$. We also notice that the number $3$ appears in both the sets $A$ and $B$. In other words, $n$ $(A \cap B)$ = $1$.
The required probability is $P(A \cup B)$.  Now let us calculate the probabilities which are obvious. 

$P(A)$ = $[\frac{n(A)}{n(S)}]$ = $\frac{3}{6}$ = $\frac{1}{2}$, $P(B)$ = $[\frac{n(B)}{n(S)}]$ = $\frac{2}{6}$ = $\frac{1}{3}$ and $P(A \cap B)$ = $[\frac{n(A \cap B)}{n(S)}]$ = $\frac{1}{6}$

Therefore, $P(A \cup B)$ = $(\frac{1}{2})$ + $(\frac{1}{3})$ - $(\frac{1}{6})$ = $(\frac{4}{6})$ = $(\frac{2}{3})$

We can check this by finding out the probability $P(A \cup B)$ directly. If $A$ is $\{2, 3, 5\}$ and $B$ is $\{3, 6\}$, then $A( \cup B)$ = $\{2, 3, 5, 6\}$ means $n(A \cup B)$ = $4$. Hence $P(A \cup B)$ = $[\frac{n(A \cup B)}{n(S)}]$ = $\frac{4}{6}$ = $\frac{2}{3}$, which is the same answer. Then why do we have to use the formula $P\ (A \cup B)$ = $P(A) + P(B) - P(A \cap B)$ instead of working out directly? Well, in this particular problem the direct approach may be easier but may not be so in all the cases. To list the events of union of different sets may be difficult if the elements are large in number and also if there are more than $2$ sets.
Example 2: 

From a well shuffled pack of standard cards a card is picked up randomly. What is the probability that the card picked up is a face card or a diamond?


Let $A$ be set of face cards and since there are $16$ face cards in a standard deck of cards, $n (A)$ = $16$. Let $B$ be the set of cards of diamond suit and since there are $13$ cards in each suit, $n(B)$ = $13$. And $n(S)$ = $52$, since the total number of cards is $52$. In the $13$ cards of diamond suit, there are $4$ face cards which are also the part of $A$. Therefore $n(A \cap B)$ = $4$. 

Now to answer the given question, we need to find $P (A \cup B)$. In this case the reader may realize that it is cumbersome to list out the elements for $A \cup B$. But it is easier to answer the question by,

$P (A \cup B)$ = $P(A) + P(B) - P(A \cap B)$ = $(\frac{16}{52})$ + $(\frac{13}{52})$ - $(\frac{4}{52})$ = $\frac{25}{52}$.
Example 3: 

If a fair die is rolled, what is the probability of rolling on an odd number or a multiple of $4$?


Let $A$ be set of events of rolling on an odd number $1, 2$ or $3$. So $n(A)$ = $3$. Let $A$ be set of events of rolling on a multiple of $4$. There is only one possibility here because $4$ is the only number in the sample space.  So, $n (B)$ = $1$ and $n(S)$ = $6$

In this case, the occurrence of both the events rolling on an odd number and rolling on a multiple of $4$ is not possible and hence $A$ and $B$ are mutually exclusive events. Therefore, $P(A \cap B)$ = $0$. So,  

$P (A \cup B)$ = $P(A) + P(B)$ = $(\frac{3}{6})$ + $(\frac{1}{6})$ = $\frac{4}{6}$ = $\frac{2}{3}$.
Example 4: 

A charity organization awards scholarships for three different fields, $20$ in mathematics, $15$ in chemistry and $10$ in physics. A student is entitled for scholarship for more than one field. If total $28$ students were awarded the scholarships and only two students got it for all the three fields, how many received scholarships in exactly two of the three fields?


Let $A, B$ and $C$ represents the fields of math, chemistry and physics respectively. Given that, $n (A \cup B \cup C)$ = $28,\ n(A)$ = $20$, $n(B)$ = $15$, $n(C)$ = $10$ and $n\ (A \cap B \cap C)$ = $2$. Let $'x',\ 'y'$ and $'z'$ are the students awarded only for $\frac{math}{chemistry}$, $\frac{chemistry}{physics}$ and $\frac{physics}{math}$ respectively.

From the formula,

$n (A \cup B \cup C)$ = $n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (C \cap A) + n ( A \cap B \cap C),\ 28$ = $20 + 15 + 10 - [n (A \cap B ) + n (B \cap C ) + n (C \cap A )] + 2$ which gives $n (A \cap B ) + n (B \cap C ) + n (C \cap A )$ = $19$

Now, obviously $n (A \cap B)$ = $x + 2,\ n (B \cap C)$ = $y + 2$ and $n (C \cap A)$ =  $z + 2$.

Therefore, $(x + 2) + (y + 2) + (z + 2)$ = $19$ or $x + y + z$ = $13$.

Thus, $13$ students were awarded scholarships in exactly two fields.