It is estimated that 0.5\% of the population of a city has certa

Monincbh

Monincbh

Answered question

2021-11-12

It is estimated that 0.5% of the population of a city has certain disease. A diagnostic test has a probability 0.95 of giving a positive result when applied to a person has the disease, and a probability 0.10 of giving a false positive result. Suppose that the test is now done to a person from the city.
Calculate the following probabilities:
(a) The probability that the test result will be positive;
(b) The probability that given a positive result, the person has the disaese;
(c) The probability that given a negative result, the person does not have the disease;
(d) The probability that the person will be misclassified

Answer & Explanation

Anot1954

Anot1954

Beginner2021-11-13Added 16 answers

Step 1
a) Obtain the probability that test will be positive.
The probability that test will be positive is obtained below as follows:
From the information, given that
Let P denote the event that the person has tested with positive result.
Let N denote the event that the person has tested with negative result.
Let D denote the event that the person has disease. That is, P(D)=0.005.
Let ND denote the event that person do not have disease. P(ND)=0.995(=10.005)
Also given that
P(PD)=0.95
{(ND)=0.05(=10.95)
P(PND)=0.10
P(NND)=0.90(10.10)
The required probabiluty is,
P(P)=[P(PD)]+[P(PND)]
=[P(PD)×P(D)]+[P(PND)×P(ND)]
=[0.95×0.005]+[0.10×0.995]
=0.00475+0.0995
=0.10425
0.1043
The probability that test will be positive is 0.1043.
Step 2
b) Obtain the probability that given a positive result, the person has the disease.
The probability that given a positive result, the person has the disease is obtained below as follows:
The required probability is,
P(DP)=P(PD)×P(D)P(P)[P(DP)=P(PD)×P(D)]
=0.95×0.0050.10425
=0.004750.10425
0.0456
The probability that given a positive result, the person has the disease is 0.0456.
Step 3
c ) Obtain the probability that given a negative result, the person does not have the disease.
The probability that given a negative result, the person does not have the disease is obtained below as follows:
The required probability is,
P(NDN)=P(NND)×P(ND)P(N)[P(NND)=P(NND)×P(ND)]
=0.90×0.99510.10425[P(N)=1P(P)]
=0.89550.89575
=0.9997
The probability that given a negative result, the person does not have the disease is 0.9997.

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