# 2.1.3.2.3 - Unions

2.1.3.2.3 - UnionsA **union** is communicated using the symbol ∪. \(P(A \cup B)\) is read as "the probability of A or B." Note that in mathematics, "or" means "and/or." The Venn diagram below depicts the union of A and B.

If the values of P(A), P(B), and P(A ∩ B) are all known, the formula below can be used to compute the union of A and B. Conceptually, the union of A and B is equal to A plus B minus the overlap of A and B.

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

## Example: Hearts or Spades

What is the probability of randomly selecting a card from a standard 52-card deck that is a heart or spade?

In a standard 52-card deck, 13 cards are hearts, 13 cards are spades, and no cards are both a heart and a spade.

\(P(heart) = \dfrac{13}{52}\)

\(P(spade) = \dfrac{13}{52}\)

\(P(heart \cap spade) = \dfrac{0}{52}\)

Using the formula given above:

\(P(heart \cup spade)=\dfrac{13}{52}+\dfrac{13}{52}-\dfrac{0}{52}= \dfrac {26}{52}=0.5\)

## Example: Hearts or Aces

What is the probability of randomly selecting a card from a standard 52-card deck that is a heart or an ace?

In a standard 52-card deck, 13 cards are hearts and 4 cards are aces. There is one ace of hearts.

\(P(heart) = \dfrac{13}{52}\)

\(P(ace) = \dfrac{4}{52}\)

\(P(heart \cap ace) = \dfrac{1}{52}\)

Using the formula given above:

\(P(heart \cup ace)=\dfrac{13}{52}+\dfrac{4}{52}-\dfrac{1}{52}=\dfrac{16}{52}=0.308\)

## Example: Part-Time or World Campus

The two-way contingency table below displays the Penn State's undergraduate enrollments from Fall 2019 in terms of status (full-time and part-time) and primary campus (data from the Penn State Factbook).

Full-Time | Part-Time | Total | |
---|---|---|---|

University Park | 39529 | 1110 | 40639 |

Commonwealth Campuses | 24306 | 2794 | 27100 |

PA College of Technology | 4110 | 871 | 4981 |

World Campus | 2574 | 5786 | 8360 |

Total | 70519 | 10561 | 81080 |

What proportion of all students are part-time students or World Campus students?

When we have a contingency table, we can take the appropriate values from the table as opposed to using the formula given above. In this table there are 1110 part-time University Park students, 2794 part-time Commonwealth Campus students, 871 PA College of Technology students, 5786 part-time World Campus students, and 2574 full-time World Campus students. Combined, these are all of the cells this question is asking about.

\(P(PartTime \cup WorldCampus)=\dfrac{1110+2794+871+5786+2574}{81080}=\dfrac{13135}{12242}=0.162\)

Note that the final answer would be the same if we had used the formula:

\(P(PartTime \cup WorldCampus) = \dfrac{10561}{81080}+\dfrac{8360}{81080}-\dfrac{5786}{81080}= \dfrac{13135}{81080}=0.162\)