The probability that if a malpractice suit is filed, it originated in hospital 1 = P(B1|A) =
To solve this problem we will use Baye's theorem which states
P(B1 |A) = P(A/B1)P(B1) / P(A|B1)P(B1) + P(A|B2)P(B2) + P(A|B3)P(B3) (1)
where P(A|B) is the probability of event A occurring given that event B has already taken place.
Probability of hospital 1 having the case = P(B1) = 50
Probability of hospital 2 having the case = P(B2) = 30
Probability of hospital 3 having the case = P(B3) = 20
Probability of case resulting in malpractice in hospital 1= P(A/B1) =0.001
Probability of case resulting in malpractice in hospital 2= P(A/B2) =0.005
Probability of case resulting in malpractice in hospital 3= P(A/B3) =0.008
The probability that if a malpractice suit is filed, it originated in hospital 1 = P(B1|A)
Using equation (1)
P(B1|A) = (0.001)(50) / (0.001)(50) + (0.005)(30) + (0.008)(20)
= 0.05 / ( 0.05 + 0.15 + 0.16 )
= 0.05 / 0.36
= 0.138
Problem based on Bayes's theorem
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