Question

One in a thousand people are afflicted with the dreaded zigma disease. A pharmaceutical company develops a test for this disease and brags that only 5% of those with the disease test negative for it and 1% of those who do not have zigma will test positive for it.

What is the probability that a person who tests positive actually has zigma?

a. 9.99%
b. 1.09%
c. 8.68%
d. 5.01%
e. 9.51%

Answer 1

Using Bayesian probability, the probability that a person who tests positive for zigma actually has the disease is calculated as approximately 8.68%.

Step-by-step explanation:

To find the probability that a person who tests positive for zigma actually has the disease, we need to use Bayesian probability. We'll use the given percentages to calculate this probability, a formula often used for these problems is:

P(A|B) = P(B|A) * P(A) / [P(B|A) * P(A) + P(B|~A) * P(~A)],

where:

• P(A) is the probability of having the disease (0.1% or 0.001).
• P(B|A) is the probability of testing positive given that one has the disease, which is 95% (since only 5% test negative).
• P(~A) is the probability of not having the disease (99.9% or 0.999).
• P(B|~A) is the probability of testing positive given that one does not have the disease (1%).

Let's do the calculations:

P(A|B) = (0.95 * 0.001) / (0.95 * 0.001 + 0.01 * 0.999),

P(A|B) = 0.00095 / (0.00095 + 0.00999),

P(A|B) = 0.00095 / 0.01094,

P(A|B) ≈ 0.0868 or 8.68%.

Therefore, the probability that a person who tests positive actually has zigma is roughly 8.68%.