Insurance status—covered ( C ) or not covered ( N )—is determined for each indiv

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 O1 = (C, C), O2 = (C, N), O3 = (N, C), and O4 = (N, N). Suppose that probabilities are P(O1) = 0.81, P(O2) = 0.09, P(O3) = 0.09, and P(O4) = 0.01.

(a) What outcomes are contained in A, the event that at most one patient is covered?

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

A = {(C, C), (N, C), (N, N)}    

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

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

A = {(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)}

B = {(C, C), (N, N)}    

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

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

the empty set

What is P(B)?
P(B) =

Explanation / Answer

a. outcomes are contained in A, the event that at most one patient is covered:

Maximum 1 person should be covered:

So A = {(C, N), (N, C), (N, N)}

As both cannot be C

P(A) = P(O2) + P(O3) + P(O4) = 0.09 + 0.09 + 0.01 = 0.19

b. outcomes are contained in B, the event that the two patients have the same status with respect to coverage:

B = {(C, C), (N, N)}    

As both have to be C or both have to be N

P(B) = P(O1) + P(O4) = 0.81+ 0.01 = 0.82

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