Question

A health company develops a testing device to test for the coronavirus disease. A specific test with the device has a false positive rate of 1% and a false negative rate of 2%. Suppose that 5% of all the people who go for the test are infected by coronavirus, and we randomly select one person, given that the person tests positive, what is the probability that he is really coronavirus positive? (Leave your answer in 3 decimal places)

Answer 1

Using conditional probability, it is found that there is a 0.838 = 83.8% probability that he is really coronavirus positive.

What is Conditional Probability?

Conditional probability is the probability of one event happening, considering a previous event. The formula is:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which:

• P(B|A) is the probability of event B happening, given that A happened.

• $P(A \cap B)$ is the probability of both A and B happening.

• P(A) is the probability of A happening.

For this problem, the events are given as follows:

• Event A: Positive test.

• Event B: Person has the disease.

The percentages associated with a positive test is:

• 98% of 5%(person has the disease).

• 1% of 95%(person does not have the disease).

Hence:

P(A) = 0.98 x 0.05 + 0.01 x 0.95 = 0.0585.

The probability of both a positive test and having the disease is:

$P(A \cap B) = 0.98 \times 0.05 = 0.049$

Hence the conditional probability is:

P(B|A) = 0.049/0.0585 = 0.838.

0.838 = 83.8% probability that he is really coronavirus positive.

More can be learned about conditional probability at brainly.com/question/14398287

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