Question

2% of the population suffers from the terrible disease conditionitis. The test for conditionitis comes back positive 98% of the time when taken by those who have the disease; it comes back negative 77% of the time when taken by those who do not have the disease.

If a randomly chosen person is given the test and the test comes back positive for conditionitis, what is the probability that he or she actually has the disease?

Answer 1

The probability that a randomly chosen person actually has the disease given that the test comes back positive is 8%.

Step-by-step explanation:

To find the probability that a randomly chosen person actually has the disease given that the test comes back positive, we can use Bayes' theorem. Let D represent the event of having the disease, and T represent the event of testing positive for the disease.

The probability of having the disease, P(D), is 2% or 0.02.

The probability of testing positive given that the person has the disease, P(T|D), is 98% or 0.98.

The probability of testing positive given that the person does not have the disease, P(T|not D), is 1 minus the probability of testing negative, which is 100% - 77% or 23% or 0.23.

Using these probabilities, we can calculate the probability of actually having the disease given that the test comes back positive using Bayes' theorem:

P(D|T) = (P(T|D) * P(D)) / ((P(T|D) * P(D)) + (P(T|not D) * P(not D)))

P(D|T) = (0.98 * 0.02) / ((0.98 * 0.02) + (0.23 * 0.98)) = 0.08

Therefore, the probability that a randomly chosen person actually has the disease given that the test comes back positive is 8% or 0.08.

Answer 2

8% probability that he or she actually has the disease

Step-by-step explanation:

We use the Bayes Theorem to solve this question.

Bayes Theorem:

Two events, A and B.

`P(B|A) = (P(B)*P(A|B))/(P(A))`

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

If a randomly chosen person is given the test and the test comes back positive for conditionitis, what is the probability that he or she actually has the disease?

This means that:

Event A: Test comes back positive.

Event B: Having the disease.

Test coming back positive:

2% have the disease(meaning that P(B) = 0.02), and for those, the test comes positive 98% of the time. This means that
`P(A|B) = 0.98`

For the 100-2 = 98% who do not have the disease, the test comes back positive 100-77 = 23% of the time.

Then

`P(A) = 0.02*0.98 + 0.98*0.23 = 0.245`

Finally:

`P(B|A) = (0.02*0.98)/(0.245) = 0.08`

8% probability that he or she actually has the disease