Question

A family consisting of three people- \(\mathrm{P}_{1}, \mathrm{P}_{2}\), and \(\mathrm{P}_{3}\) \- belongs to a medical clinic that always has a physician at each of stations 1,2, and 3 . During a certain week, each member of the family visits the clinic exactly once and is randomly assigned to a station. One experimental outcome is \((1,2,1)\), which means that \(\mathrm{P}_{1}\) is assigned to station \(1 .\) \(\mathrm{P}_{2}\) to station 2, and \(\mathrm{P}_{3}\) to station 1 a. List the 27 possible outcomes. (Hint: First list the nine outcomes in which \(\mathrm{P}_{1}\) goes to station 1 , then the nine in which \(\mathrm{P}_{1}\) goes to station 2 , and finally the nine in which \(\mathrm{P}_{1}\) goes to station 3 ; a tree diagram might help.) b. List all outcomes in the event \(A\), that all three people go to the same station. c. List all outcomes in the event \(B\), that all three people go to different stations. d. List all outcomes in the event \(C\), that no one goes to station \(2 .\) e. Identify outcomes in each of the following events: \(B^{C}\), \(C^{C}, A \cup B, A \cap B, A \cap C\)

Short answer

Event \(A\) contains the outcomes: (1,1,1), (2,2,2), (3,3,3). Event \(B\) contains: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1). Event \(C\) contains: (1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (3,3,3). Event \(B^C\) is all outcomes minus \(B\), event \(C^c\) is all outcomes minus \(C\), event \(A \cup B\) are all outcomes in \(A\) and \(B\) without duplicates, event \(A \cap B\) is null (no common outcomes), and \(A \cap C\) is (1,1,1), (3,3,3).

Step-by-step solution

List all outcomes

To list all possible outcomes, start by considering where \(\mathrm{P}_{1}\) can go. For each choice of \(\mathrm{P}_{1}\), there are 3 possibilities for \(\mathrm{P}_{2}\), \(\mathrm{P}_{3}\), creating 3 * 3 = 9 outcomes per choice of \(\mathrm{P}_{1}\). Therefore, there are 27 total outcomes, broken down as follows:\n\nFor \(\mathrm{P}_{1}\) at station 1: (1,1,1), (1,1,2), (1,1,3), (1,2,1), (1,2,2), (1,2,3), (1,3,1), (1,3,2), (1,3,3).\nFor \(\mathrm{P}_{1}\) at station 2: (2,1,1), (2,1,2), (2,1,3), (2,2,1), (2,2,2), (2,2,3), (2,3,1), (2,3,2), (2,3,3).\nFor \(\mathrm{P}_{1}\) at station 3: (3,1,1), (3,1,2), (3,1,3), (3,2,1), (3,2,2), (3,2,3), (3,3,1), (3,3,2), (3,3,3).

Identify Event A

Event \(A\) is that all three people go to the same station. This event includes the following outcomes: (1,1,1), (2,2,2), (3,3,3).

Identify Event B

Event \(B\) is that all three people go to different stations. This event includes the following outcomes: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1).

Identify Event C

Event \(C\) is that no one goes to station 2. This event includes the following outcomes: (1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (3,3,3).

Identify Events Compositions

Here are the outcomes for other events asked:\n\nEvent \(B^c\), i.e., not \(B\): Includes all outcomes that are not in \(B\). This is all outcomes minus the outcomes of \(B\). You simply look at the list of all outcomes and exclude the ones that are listed under \(B\).\n\nEvent \(C^c\), i.e., not \(C\): Includes all outcomes that are not in \(C\). This is all outcomes minus the outcomes of \(C\). You simply look at the list of all outcomes and exclude the ones that are listed under \(C\).\n\nEvent \(A \cup B\) (the union of \(A\) and \(B\)): Includes all outcomes that are in \(A\), in \(B\), or in both. This set can be found by combining the outcome lists for \(A\) and for \(B\), and removing duplicates.\n\nEvent \(A \cap B\) (the intersection of \(A\) and \(B\)): Includes all outcomes that are both in \(A\) and \(B\). It can be found by taking the list of outcomes in \(A\) and removing all that are not in \(B\). As \(A\) and \(B\) do not share any outcomes, this set is empty.\n\nEvent \(A \cap C\) (the intersection of \(A\) and \(C\)): Includes all outcomes that are both in \(A\) and \(C\). It can be found by taking the list of outcomes in \(A\) and removing all that are not in \(C\). The intersection is (1,1,1), (3,3,3).