Use Bayes' theorem to solve this problem.
A certain virus infects 10 in every 5000 people. A test used to detect the virus in a person is positive 96% of the time if the person has the virus and 2% of the time if the person does not have the virus (false positive).
Find the following probabilities. Enter your answer as a percent, rounded to the nearest HUNDREDTH of a percent.
(a) The probability that a person really has the virus given that they have tested positive
P(virus| +) = Answer
%
(b) Find the probability that a person does not have the virus given that they test negative
P(no virus | −) =Answer
%
To solve this problem using Bayes' theorem, we need to understand the given probabilities and calculate the desired probabilities step by step.
Let's define the following events:
Virus: A person has the virus.
No Virus: A person does not have the virus.
Positive: A person tests positive for the virus.
Negative: A person tests negative for the virus.
We are given the following probabilities:
P(Virus) = 10/5000 = 0.002 (probability of having the virus)
P(No Virus) = 1 - P(Virus) = 1 - 0.002 = 0.998 (probability of not having the virus)
P(Positive | Virus) = 0.96 (probability of testing positive if the person has the virus)
P(Negative | Virus) = 1 - P(Positive | Virus) = 0.04 (probability of testing negative if the person has the virus)
P(Positive | No Virus) = 0.02 (probability of testing positive if the person does not have the virus)
P(Negative | No Virus) = 1 - P(Positive | No Virus) = 0.98 (probability of testing negative if the person does not have the virus)
Now, we can calculate the desired probabilities using Bayes' theorem:
(a) The probability that a person really has the virus given that they have tested positive:
P(Virus | Positive) = (P(Positive | Virus) * P(Virus)) / [P(Positive | Virus) * P(Virus) + P(Positive | No Virus) * P(No Virus)]
P(Virus | Positive) = (0.96 * 0.002) / [(0.96 * 0.002) + (0.02 * 0.998)]
P(Virus | Positive) = 0.00192 / (0.00192 + 0.01996)
P(Virus | Positive) = 0.00192 / 0.02188
P(Virus | Positive) ≈ 0.0879 → 8.79%
Therefore, the probability that a person really has the virus given that they have tested positive is approximately 8.79%.
(b) The probability that a person does not have the virus given that they test negative:
P(No Virus | Negative) = (P(Negative | No Virus) * P(No Virus)) / [P(Negative | Virus) * P(Virus) + P(Negative | No Virus) * P(No Virus)]
P(No Virus | Negative) = (0.98 * 0.998) / [(0.04 * 0.002) + (0.98 * 0.998)]
P(No Virus | Negative) = 0.977004 / (0.00008 + 0.979844)
P(No Virus | Negative) = 0.977004 / 0.979924
P(No Virus | Negative) ≈ 0.9969 → 99.69%
Therefore, the probability that a person does not have the virus given that they test negative is approximately 99.69%.