## 5.2 Probability of a Union

For any two events \(A\) and \(B\), we have the probability of a union (also known as the addition rule) which states that the probability of \(A\) or \(B\) occurring can be written as: \[Pr(A \cup B)=P(A)+P(B)-P(A \cap B)\] If \(Pr(A \cap B)=0\), then we call them disjoint events.

For example, suppose that researchers are interested in the substances found in people’s blood. Let \(A=\{Alcohol\}\), \(B=\{Cocaine \}\), and \(A \cap B = \{Both \}\). Assume that the probabilities are as follows: \[Pr(A)=0.86\\ Pr(B)=0.35\\ Pr(A \cap B)=0.29\] Hence, \(Pr(A \cup B)\) is calculated as follows: \[Pr(A \cup B)= 0.86+0.35-0.29 = 0.92\] If you are intrigued by this example on how to measure drug consumption, check out this EU Project.

For a second example, assume the following events taken from the police by controlling semi-trucks: \(A=\{\text{faulty breaks}\}\), \(B=\{\text{bad tires}\}\), and \(A \cup B =\{\text{faulty breaks and/or bad tires}\}\). Let the probabilities be as follows: \[Pr(A)=0.23\\ Pr(B)=0.24\\ Pr(A \cap B)=0.09\] Using the above equation, we can determine that \(P(A \cup B)=0.23+0.24-0.09=0.38\). Note that if the events are mutually exclusive, the term \(Pr(A \cap B)\) is equal to 0.

Lastly, consider the data below on the gate arrival of airplanes during a week at a mid-sized airport. Everything not within +/- 10 minutes is considered not on time. \(Flights\) represents the total number of arriving flights which is 275.

Arrival | Event | Flights | Probability |
---|---|---|---|

Less than 10 minutes early | A | 55 | 0.20 |

Within +/- 10 minutes | B | 121 | 0.44 |

More than 10 minutes late | C | 99 | 0.36 |

The probability of a flight not begin on time is calculated as follows: \[Pr(A \cup C)=0.2+0.36=0.56\]