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.

Answer 1

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.