This is a statistics problem.

A laboratory test for the detection of a certain disease give a positive result 5 percent of the time for people who do not have the disease. The test gives a negative result 0.3 percent of the time for people who have the disease. Large-scale studies have shown that the disease occurs in about 2 percent of the population.

What is the probability that a person selected at random would test positive for this disease?
For this one, I got 0.0194, so 1.94% but I wasn’t sure if the answer is correct.

What is the probability that a person selected at random who test positive for the disease does not have the disease?
This one was confusing to me… I got 0.71637, or 71.637% but I doubt this is right. Please help!

Consider the four possibilities and their probabilities.

has disease and tests positive:
(0.02)x(0.997)= 0.0199
has disease but tests negative:
(0.02)x(0.003)= 0.0001
has no disease and tests negative:
(0.98)x(0.95) = 0.9310
has no disease but tests positive:
(0.98)x(0.05) = 0.0490
Note the probabilities they add up to 1.000, as they should.
The answer to the first question (positive test probability) is
0.0199 + 0.0490 = 0.0689

The answer to the second question is:
0.0490/(0.0199+0.0490)= 0.7112

This is a rather large ratio of "false positives" and comes about because the disease is rare (2% of population) and there are many more false positives than "true" positives

That makes perfect sense. Thank you very much.

hey, we just did this problem in class today...this is from the ap stats 1197 free response prep...the first question's ans. is 0.06894 and the second ques.'s ans. is 0.710763

To solve these statistics problems, we can use conditional probability and Bayes' theorem. Let's break down each question step by step.

Question 1: What is the probability that a person selected at random would test positive for this disease?

To find the probability that a person tests positive for the disease, you need to consider two cases:
1) The person has the disease and tests positive.
2) The person does not have the disease and still tests positive.

Let's calculate the probability of each case separately and then add them up.

Case 1: The person has the disease and tests positive.
From the problem statement, we are given that the disease occurs in about 2 percent of the population. Therefore, the probability of randomly selecting someone with the disease is 0.02 (2 percent).

Given that the test gives a negative result 0.3 percent of the time for people who actually have the disease, the probability of the person testing positive, given that they have the disease, is 1 - 0.003 = 0.997.

So, the probability of case 1 is:
P(case 1) = Probability of having the disease × Probability of testing positive given the disease = 0.02 × 0.997 = 0.01994

Case 2: The person does not have the disease and tests positive.
We are given that the test gives a positive result 5 percent of the time for people who do not have the disease. So, the probability of a person not having the disease and testing positive is 0.05 (5 percent) since this is the false positive rate.

Therefore, the probability of case 2 is:
P(case 2) = Probability of not having the disease × Probability of testing positive given no disease = 0.98 × 0.05 = 0.049

Now, let's add the probabilities of both cases to find the total probability that a person selected at random would test positive for this disease:

Total probability = P(case 1) + P(case 2)
= 0.01994 + 0.049
= 0.06894

So, the probability that a person selected at random would test positive for this disease is approximately 0.069, or 6.9%.

Question 2: What is the probability that a person selected at random who tests positive for the disease does not have the disease?

To find this probability, we need to use Bayes' theorem. The formula for Bayes' theorem is:

P(A|B) = (P(B|A) × P(A)) / P(B)

where:
P(A|B) is the probability of event A occurring given that event B has already occurred.
P(B|A) is the probability of event B occurring given that event A has already occurred.
P(A) is the probability of event A occurring.
P(B) is the probability of event B occurring.

In this case:
A = Does not have the disease
B = Tests positive for the disease

Let's calculate the probability that a person selected at random who tests positive for the disease does not have the disease:

P(A|B) = (P(B|A) × P(A)) / P(B)

P(A|B) = (Probability of testing positive given no disease × Probability of not having the disease) / Probability of testing positive

From case 2 in question 1, we know that the probability of testing positive given no disease is 0.05 (5 percent), and the probability of not having the disease is 0.98.

The probability of testing positive is the total probability from question 1, which we found to be 0.06894.

Now, let's plug in the values and calculate:

P(A|B) = (0.05 × 0.98) / 0.06894 = 0.7095

So, the probability that a person selected at random who tests positive for the disease does not have the disease is approximately 0.709, or 70.9%.

I hope this explanation helps you understand how to solve these statistics problems!