Question

A test is used to determine if people have a predisposition for the formation of a blood clot inside a blood vessel that obstructs the flow of blood through the circulatory system. It is believed that 3% of people actually have this predisposition. The test is 99% accurate if a person actually has the predisposition, meaning that the probability of a positive test result when a person actually has the predisposition is 0.99. The test is 98% accurate if a person does not have the predisposition. What is the probability that a randomly selected person who tests positive for the predisposition by the test actually has the predisposition

Answer 1

Answer:

The probability that a randomly selected person who tests positive for the predisposition by the test actually has the predisposition is 0.6049.

Step-by-step explanation:

We are given that it is believed that 3% of people actually have this predisposition. The test is 99% accurate if a person actually has a predisposition.

The test is 98% accurate if a person does not have a predisposition.

Let the probability that people actually have predisposition = P(PD) = 0.03

The probability that people do not have a predisposition = P(PD') = 1 - P(PD) = 1 - 0.03 = 0.97

Let A = event that the test is accurate

So, the probability that the test is accurate if a person actually has a predisposition = P(A/PD) = 0.99

The probability that the test is correct if a person actually has a predisposition = P(A/PD') = 1 - 0.98 = 0.02

Now,  the probability that a randomly selected person who tests positive for the predisposition by the test actually has the predisposition = P(PD/A)

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

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

                    = $\frac{0.03 \times 0.99}{0.03 \times 0.99+0.97 \times 0.02}$

                    = $\frac{0.0297}{0.0491}$ = 0.6049.