Part 1 of 2 Points: 0 of 1 Save A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive \( 80 \% \) of the time when the person has the virus and \( 15 \% \) of the time when the person does not have the virus. (This \( 15 \% \) result is called a false positive.) Let A be the event "the person is infected" and B be the event "the person tests positive." (a) Using Bayes' Theorem, when a person tests positive, determine the probability that the person is infected. (b) Using Bayes' Theorem, when a person tests negative, determine the probability that the person is not infected. (i) Click the icon to the right to review Bayes' Theorem. (a) The probability that a person is infected when a person tests positive is \( \square \) (Do not round until final answer. Then round to three decimal places as needed.)

To find the probability that a person is infected given a positive test result (P(A | B)), we can use Bayes' Theorem: \[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \] Where: - P(A) is the prior probability that a person is infected (1 in 300). - P(B | A) is the probability of testing positive if the person is infected (80%). - P(B) is the overall probability of testing positive, which can be calculated using the law of total probability. **Step 1: Calculate the probabilities** 1. **P(A)**: The probability of being infected is: \[ P(A) = \frac{1}{300} \approx 0.00333 \] 2. **P(B | A)**: The probability of testing positive given the person is infected is: \[ P(B | A) = 0.8 \] 3. **P(B | A')**: The probability of testing positive given that the person is not infected. The probability of not being infected (A') is: \[ P(A') = 1 - P(A) = 1 - \frac{1}{300} \approx 0.99667 \] Thus, the probability of testing positive when not infected (false positive) is: \[ P(B | A') = 0.15 \] 4. **P(B)**: We can calculate P(B) using the law of total probability: \[ P(B) = P(B | A) \cdot P(A) + P(B | A') \cdot P(A') \] \[ P(B) = (0.8 \cdot 0.00333) + (0.15 \cdot 0.99667) \] \[ P(B) \approx 0.002664 + 0.1495 \approx 0.152164 \] **Step 2: Substitute into Bayes' Theorem** Now we can find P(A | B): \[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} = \frac{0.8 \cdot 0.00333}{0.152164} \approx \frac{0.002664}{0.152164} \approx 0.0175 \] Thus, rounding to three decimal places, the probability that a person is infected when they test positive is: \[ \boxed{0.018} \] --- For part (b), now we want to find the probability that a person is NOT infected given a negative test result (P(A' | B')). Again, we can apply Bayes' Theorem: \[ P(A' | B') = \frac{P(B' | A') \cdot P(A')}{P(B')} \] Where: - P(B' | A') is the probability of testing negative if the person is not infected, which is 1 - P(B | A') = 1 - 0.15 = 0.85. - P(B') is the overall probability of testing negative, which we can calculate similarly. **Step 1: Calculate P(B')** Using \[ P(B') = P(B' | A) \cdot P(A) + P(B' | A') \cdot P(A') \] And knowing P(B' | A) = 1 - P(B | A) = 0.2: \[ P(B') = (0.2 \cdot 0.00333) + (0.85 \cdot 0.99667) \] \[ P(B') \approx 0.000666 + 0.847195 \approx 0.847861 \] **Step 2: Substitute into Bayes' Theorem** Now we can find P(A' | B'): \[ P(A' | B') = \frac{P(B' | A') \cdot P(A')}{P(B')} = \frac{0.85 \cdot 0.99667}{0.847861} \approx \frac{0.848169}{0.847861} \approx 1.0003 \] So the probability that a person is not infected when they test negative (rounded to three decimal places) is: \[ \boxed{1.000} \] It suggests that there’s a very high probability they'll be not infected if they test negative, matching with our intuition given the effectiveness of the test in this scenario!