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

Answer:

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.

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

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:

$P(B|A) = \frac{0.02*0.98}{0.245} = 0.08$

8% probability that he or she actually has the disease