A blood test indicates the presence of a particular disease 95% of the time when the disease is actually present. The same test indicates th

Question

A blood test indicates the presence of a particular disease 95% of the time when the disease is actually present. The same test indicates the presence of the disease 0.5% of the time when the disease is not actu- ally present. One percent of the population actually has the disease. Calculate the probability that a person actually has the disease given that the test indicates the presence of the disease.

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Josephine 2 hours 2021-09-15T19:23:37+00:00 1 Answer 0

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    2021-09-15T19:25:05+00:00

    Answer:

    Probability that a person actually has the disease given that the test indicates the presence of the disease is 0.657.

    Step-by-step explanation:

    We are given that a blood test indicates the presence of a particular disease 95% of the time when the disease is actually present.

    The same test indicates the presence of the disease 0.5% of the time when the disease is not actually present. One percent of the population actually has the disease.

    Let the Probability that person actually has the disease = P(A) = 0.01

    Probability that person actually doesn’t has the disease = P(A’) = 1 – P(A) = 1 – 0.01 = 0.99

    Also, let PD = event that there is a presence of the disease

    So, Probability that test indicates the presence of the disease given the fact that the disease is actually present = P(PD/A) = 0.95

    Probability that test indicates the presence of the disease given the fact that the disease is not actually present = P(PD/A’) = 0.005

    Now, the probability that a person actually has the disease given that the test indicates the presence of the disease = P(A/PD)

    We will use Bayes’ theorem to calculate the above probability.

    SO,   P(A/PD)  =  \frac{P(A) \times P(PD/A)}{P(A) \times P(PD/A) + P(A')\times P(PD/A')}

                           =  \frac{0.01 \times 0.95}{0.01 \times 0.95+ 0.99\times 0.005}      

                           =  \frac{190}{289}  =  0.657

    Hence, the required probability is 0.657.

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