Question

Insurance status - covered (C) or not covered (N) \- is determined for each individual arriving for treatment at a hospital's emergency room. Consider the chance experiment in which this determination is made for two randomly selected patients. The simple events are \(O_{1}=(\mathrm{C}, \mathrm{C})\) \(O_{2}=(\mathrm{C}, \mathrm{N}), O_{3}=(\mathrm{N}, \mathrm{C})\), and \(O_{4}=(\mathrm{N}, \mathrm{N}) .\) Suppose that probabilities are \(P\left(O_{1}\right)=.81, P\left(O_{2}\right)=.09, P\left(O_{3}\right)=.09\), and \(P\left(O_{4}\right)=.01\). a. What outcomes are contained in \(A\), the event that at most one patient is covered, and what is \(P(A)\) ? b. What outcomes are contained in \(B\), the event that the two patients have the same status with respect to coverage, and what is \(P(B)\) ?

Short answer

For the event that at most one patient is covered, the outcomes are \(O_{2}, O_{3}, O_{4}\), and the probability is 0.19. For the event that both patients have the same coverage status, the outcomes are \(O_{1}, O_{4}\), and the probability is 0.82.

Step-by-step solution

Determine the outcomes for event A

Event A is described as 'at most one patient is covered.' This involves two simple events, where both patients are not covered, i.e. \(O_{4}=(\mathrm{N}, \mathrm{N})\) and one of the patients is covered i.e. \(O_{2}=(\mathrm{C}, \mathrm{N})\) and \(O_{3}=(\mathrm{N}, \mathrm{C})\). Therefore, \(A=\{O_{2}, O_{3}, O_{4}\}\)

Calculate the Probability of event A

The probability of event A, \(P(A)\), which is the sum of the probabilities of the simple events \(O_{2}, O_{3}, O_{4}\), i.e. \(P(A) = P(O_{2}) + P(O_{3}) + P(O_{4}) = 0.09 + 0.09 + 0.01 = 0.19.

Determine the outcomes for event B

Event B is described as 'the two patients have the same status with respect to coverage.' This involves two simple events, where both patients are not covered, i.e. \(O_{4}=(\mathrm{N}, \mathrm{N})\) or both are covered i.e. \(O_{1}=(\mathrm{C}, \mathrm{C})\). Therefore, \(B=\{O_{1}, O_{4}\}\)

Calculate the Probability of event B

The probability of event B, \(P(B)\), which is the sum of the probabilities of the simple events \(O_{1}, O_{4}\), i.e. \(P(B) = P(O_{1}) + P(O_{4}) = 0.81 + 0.01 = 0.82 .