The test to detect the presence of a liver disorder is 98% accurate for a person who has the disease and 97% accurate for a person who does not have the disease. If 3.5% of the people in a given population actually have the disorder, what is the probability that a randomly chosen person tests positive?. . 0.0343. . 0.035. . 0.06325. . 0.02895

The probability tests to detect a liver disorder:
P (presence)= 0.98 (true) 0.02 (false)
P (absence)= 0.97 (true) 0.03 (false)

P(have disorder from the population)=0.035
P(do not have disorder from the population)= 0.965

To test positive:
P[have disorder and true from P(presence)] + P[do not have disorder and false from P(absence)]

=0.035*0.98+0.965*0.03= 0.06325

The answer is 0.06325

Answer Link

saimaparvezVT saimaparvezVT · Answer 2 · 2022-06-06T14:43:17+02:00

The probability that a randomly chosen person tests positive is 0.06325 to be infected.

What is the population?

The population is the structure of the age and structure of the family and the living of the individual.

The possibility exams to come across a liver sickness:
P (positive )=0.98 (positive ) 0.02 (negative )
P (absence)= 0.97 (positive ) 0.03 (negative )\
P(have sickness from the population)=0.half P(do now no longer have sickness from the population)= 0.965
To check positive:P[have disorder and true from P(presence)] + P[do not have disorder and false from P(absence)]
=0.1/2x 0.98+0.965x0.03=0.06325
The solution is 0.06325