A test for a certain rare disease is assumed to be correct 95% of the time: if a person has the disease, the test result is positive with probability 0.95, and if the person does not have the disease, the test result is negative with probability 0.95. A person drawn at random from a certain population has probability 0.001 of having the disease.
Find the probability that a random person tests positive. (This answer will require an accuracy of 4 decimal places.)
Given that the person just tested positive, what is the probability he actually has the disease?
0.0187
To find the probability that a random person tests positive, we can use Bayes' theorem. Let's denote the event A as testing positive and event B as having the disease.
The probability of testing positive can be calculated as follows:
P(A) = P(A|B) * P(B) + P(A|not B) * P(not B)
We know that the probability of having the disease (P(B)) is 0.001, and the probability of not having the disease (P(not B)) is 1 - 0.001 = 0.999.
Given that a person has the disease, the probability of testing positive (P(A|B)) is 0.95. And given that a person does not have the disease, the probability of testing positive (P(A|not B)) is 1 - 0.95 = 0.05.
Let's substitute these values into the equation:
P(A) = (0.95 * 0.001) + (0.05 * 0.999)
P(A) = 0.00095 + 0.04995
P(A) = 0.0509
Therefore, the probability that a random person tests positive is approximately 0.0509, or 0.051 to four decimal places.
Now, to find the probability that the person actually has the disease given that they tested positive, we can use Bayes' theorem again.
Let's denote the event C as having the disease given that the person tested positive.
The probability of having the disease given a positive test result can be calculated as follows:
P(C) = P(B|A) * P(A) / P(A)
P(C) = P(B|A) * P(A) / (P(A|B) * P(B) + P(A|not B) * P(not B))
We already know the values for P(A), P(A|B), P(B), P(A|not B) and P(not B). Substituting these values:
P(C) = (0.95 * 0.001) / (0.95 * 0.001 + 0.05 * 0.999)
P(C) = 0.00095 / (0.00095 + 0.04995)
P(C) = 0.00095 / 0.0509
P(C) ≈ 0.0186
Therefore, the probability that a person actually has the disease, given that they tested positive, is approximately 0.0186, or 0.0186 to four decimal places.
To find the probability that a random person tests positive, we can use Bayes' theorem.
Let's define the following events:
A: Person has the disease
B: Person tests positive
We are given:
P(A) = 0.001 (probability that a person has the disease)
P(B|A) = 0.95 (probability of testing positive given the person has the disease)
P(B|~A) = 1 - P(B|~A) = 0.05 (probability of testing positive given the person does not have the disease)
We can now calculate the probability of a random person testing positive using the law of total probability:
P(B) = P(B|A) * P(A) + P(B|~A) * P(~A)
P(B) = 0.95 * 0.001 + 0.05 * (1 - 0.001)
P(B) = 0.00095 + 0.04995
P(B) = 0.0509 (rounded to four decimal places)
Therefore, the probability that a random person tests positive is approximately 0.0509.
Now, let's find the probability that a person actually has the disease, given that they tested positive. We can use Bayes' theorem again:
P(A|B) = P(B|A) * P(A) / P(B)
P(A|B) = 0.95 * 0.001 / 0.0509
P(A|B) = 0.00095 / 0.0509
P(A|B) ≈ 0.01864 (rounded to four decimal places)
Therefore, given that a person just tested positive, the probability that they actually have the disease is approximately 0.0186.