Insurance status-covered (C) or not covered (N s determined for each individual

Question

Insurance status-covered (C) or not covered (N s 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 C, N), O3 N, C), and O N, N). Suppose that probabilities are P(O1) 0,81, PO2) 0.09, P(O3) 0.09, and P(OA) 0.01 (a) What outcomes are contained in A, the event that at most one patient is covered? A m f(C, C), (C, N), (N, N) A (C, N), (N, C) AR (C, C), (C, N), (N, C)) What is P(A)? P(A) (b) What outcomes are contained in B, the event that the two patients have the same status with respect to coverage? B (C, C), (C, N) the empty set What is P(B)? P(B)

Explanation / Answer

Solution:

1)a) P(A) = 0.19

Outcomes for event A that atmost one patient is covered.

Atmost 1 means that either only one is covered or both of them are not covered.

A = (C, N), (N, C), (N, N)

P(A) is the probability of happening of event A.

P(A) = P(C, N) + P(N, C) + P (N, N)

P(A) = P(O2) + P(O3) + P(O4)

P(A) = 0.09 + 0.09 + 0.01

P(A) = 0.19

b) P(B) = 0.82

The outcomes for event B that two patients have same status with respect to coverage.

B = {(C,C), (N,N)}

Event B means that two patient should have same status either both covered or both uncovered.

P(B) is the probability of happening of event B.

P(B) = P(C,C)+ P(N,N)

P(B) = P(O1) + P(O4)

P(B) = 0.81 + 0.01

P(B) = 0.82

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