2.1.3.2.3 - Unions

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.

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 Section

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 Section

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 Section

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:

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