1 oV2 points | Previous Answers PODStat5 6.E.016 My Notes Ask Your Insurance sta

Question

1 oV2 points | Previous Answers PODStat5 6.E.016 My Notes Ask Your Insurance status-covered (C) or not covered (M)-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 01- (C, C), O2- (C, N), 3 (N, C),and 04(N, ). Suppose that probabilities are PrO)-0.81, PO2) - 0.09, PO30.09, and PIO4)-0.01 (a) What outcomes are contained in A, the event that at most one patient is not covered? A = {(C, C), (C, N), (N, N)} ( A = {(C, N), (N, C)} A = {(C, N), (N, C), (N, N)} What is P(A)? P(A) = 0.19 (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)) 0 the empty set What is P(B)? A(B) = 0.18 Need Help?ReadTalk to & Tutor

Explanation / Answer

Solution:

    We are given that the following events:

O1 = ( C , C ) , O2 = ( C , N ) , O3 = ( N , C ) , O4 = ( N , N)

P(O1) = 0.81

P(O2) = 0.09

P(O3) = 0.09

P(O4) = 0.01

Part a)

A = At most one patient Not Covered

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

Fifth option is correct.

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

P(A) = P(O1) + P(O2) + P(O3)

P(A) = 0.81 + 0.09 + 0.09

P(A) = 0.99

Part b)

B = The two patients have same status

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

Second option is correct.

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

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

P(B) = 0.81 + 0.01

P(B) = 0.99

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