Question

There is a new test for a new disease. For a specific version of the test, the probability that people with the disease will test positive is 0.7. The probability that someone without the disease will test positive is 0.2. Only 7% of the people taking the test actually have the disease. What is the probability that someone who tested positive actually has the disease? Report as a probability (decimal to two places).

Answer 1

The probability that someone who tested positive actually has the disease is calculated using Bayes' theorem and is found to be 0.21, or 21%.

Step-by-step explanation:

To find the probability that someone who tested positive actually has the disease, we can use Bayes' theorem. First, let's define some terms:

• P(D) is the probability of having the disease, which is 0.07 (7%).
• P(Pos|D) is the probability of testing positive given that the person has the disease, which is 0.7.
• P(Pos|Not D) is the probability of testing positive given that the person doesn't have the disease, which is 0.2.
• P(Pos) is the total probability of testing positive, which we need to calculate.

Now, we calculate P(Pos) using the total probability rule:

P(Pos) = P(D) × P(Pos|D) + P(Not D) × P(Pos|Not D)

P(Pos) = 0.07 × 0.7 + 0.93 × 0.2

P(Pos) = 0.049 + 0.186

P(Pos) = 0.235

Using Bayes' theorem, we can calculate the probability that someone who tested positive actually has the disease (P(D|Pos)) as follows:

P(D|Pos) = (P(Pos|D) × P(D)) / P(Pos)

P(D|Pos) = (0.7 × 0.07) / 0.235

P(D|Pos) = 0.049 / 0.235

P(D|Pos) = 0.21 (or 21% when expressed as a percentage)

So, the probability that someone who tested positive actually has the disease is 0.21, or 21%.