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 if the person has the virus and 5% of the time if the person does not have the virus. (This 5% 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) Find the probability that a person has the virus given that they have tested positive, i.e. find P(Has Virus | Tested Positive).

Round your answer to the nearest tenth of a percent.

P(Has Virus | Tested Positive)=     %

To solve for \( P(\text{Has Virus | Tested Positive}) \), we can use Bayes' Theorem, which states:

\[
P(\text{A | B}) = \frac{P(\text{B | A}) \cdot P(\text{A})}{P(\text{B})}
\]

where:
- \( P(\text{A}) \) is the probability that a person has the virus.
- \( P(\text{B | A}) \) is the probability that the test is positive given that the person has the virus.
- \( P(\text{B}) \) is the probability that the test is positive.

#### Given values:
1. The probability that a person has the virus, \( P(\text{A}) = \frac{1}{300} \).
2. The probability that a test is positive given the person has the virus, \( P(\text{B | A}) = 0.8 \).
3. The probability of a false positive, \( P(\text{B | Not A}) = 0.05 \).

We first need to calculate \( P(\text{B}) \), the probability that a person tests positive. This can happen in two ways:
1. The person has the virus and tests positive.
2. The person does not have the virus but still tests positive (false positive).

Thus,

\[
P(\text{B}) = P(\text{B | A}) \cdot P(\text{A}) + P(\text{B | Not A}) \cdot P(\text{Not A})
\]

where:
- \( P(\text{Not A}) = 1 - P(\text{A}) \).

### Step-by-Step Calculation

1. **Calculate \( P(\text{A}) \):**

   \[
   P(\text{A}) = \frac{1}{300} \approx 0.00333
   \]

2. **Calculate \( P(\text{Not A}) \):**

   \[
   P(\text{Not A}) = 1 - 0.00333 = 0.99667
   \]

3. **Calculate \( P(\text{B}) \):**

   \[
   P(\text{B}) = (0.8 \times 0.00333) + (0.05 \times 0.99667)
   \]
   \[
   P(\text{B}) = 0.002664 + 0.0498335 = 0.0524975
   \]

4. **Apply Bayes' Theorem to find \( P(\text{A | B}) \):**

   \[
   P(\text{A | B}) = \frac{P(\text{B | A}) \cdot P(\text{A})}{P(\text{B})}
   \]
   \[
   P(\text{A | B}) = \frac{0.8 \times 0.00333}{0.0524975}
   \]
   \[
   P(\text{A | B}) \approx \frac{0.002664}{0.0524975} \approx 0.0508
   \]

5. **Convert to a percentage and round to the nearest tenth of a percent:**

   \[
   P(\text{A | B}) \approx 5.1\%
   \]

### Final Answer:

\[
P(\text{Has Virus | Tested Positive}) = 5.1\%
\]

Would you like further details on any steps, or do you have questions?

---

Here are five related questions:
1. What is the probability of a false positive in this test scenario?
2. How does a higher accuracy in the test affect the conditional probability?
3. If the infection rate increases to 1 in 100, how does \( P(\text{Has Virus | Tested Positive}) \) change?
4. How can false negatives impact the test results?
5. What is Bayes' Theorem, and where else is it used?

**Tip:** When interpreting conditional probabilities, remember that a low prevalence rate can significantly affect the predictive value of a positive test result.

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 if the person has the virus and 5% of the time if the person does not have the virus. (This 5% 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) Find the probability that a person has the virus given that they have tested positive, i.e. find P(Has Virus | Tested Positive). Round your answer to the nearest tenth of a percent. P(Has Virus | Tested Positive) = %

Solve a virus detection probability problem using Bayes' Theorem. Learn to calculate the probability that a person has a virus given a positive test result, considering an infection rate of 1 in 300 people, a test accuracy of 80%, and a 5% false positive rate. Suitable for students in Grades 10-12.

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