# 2.1.3.2.3 - Unions

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.

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

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 39529 1110 40639 24306 2794 27100 4110 871 4981 2574 5786 8360 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:

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

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