A laboratory blood test is 95 percent effective in detecting a certain disease when
it is, in fact, present. However, the test also yields a “false positive” result for 1
percent of the healthy persons tested. (That is, if a healthy person is tested, then,
with probability .01, the test result will imply that he or she has the disease.) If .5
percent of the population actually has the disease, what is the probability that a
person has the disease given that the test result is positive?

explain me in table way 

This problem can be solved using **Bayes' Theorem**, which calculates the probability of an event given the occurrence of another event. Here, we are asked to find the probability that a person has the disease given that the test result is positive.

### Known Information:
- \( P(\text{Positive Test} \mid \text{Disease}) = 0.95 \) (Test detects disease correctly 95% of the time)
- \( P(\text{Positive Test} \mid \text{No Disease}) = 0.01 \) (False positive rate of 1%)
- \( P(\text{Disease}) = 0.005 \) (0.5% of the population has the disease)
- \( P(\text{No Disease}) = 0.995 \) (99.5% of the population does not have the disease)

We need to compute:
\[
P(\text{Disease} \mid \text{Positive Test})
\]

According to **Bayes' Theorem**:

\[
P(\text{Disease} \mid \text{Positive Test}) = \frac{P(\text{Positive Test} \mid \text{Disease}) \cdot P(\text{Disease})}{P(\text{Positive Test})}
\]

Where \( P(\text{Positive Test}) \) can be calculated as:
\[
P(\text{Positive Test}) = P(\text{Positive Test} \mid \text{Disease}) \cdot P(\text{Disease}) + P(\text{Positive Test} \mid \text{No Disease}) \cdot P(\text{No Disease})
\]

### Step-by-Step Table Representation:

| Case                          | Probability of the Case (Prior) | Test Result Probability (Conditional) | Joint Probability |
|-------------------------------|----------------------------------|---------------------------------------|-------------------|
| **Has Disease**                | \( P(\text{Disease}) = 0.005 \)  | \( P(\text{Positive} \mid \text{Disease}) = 0.95 \) | \( 0.005 \times 0.95 = 0.00475 \) |
| **No Disease (False Positive)**| \( P(\text{No Disease}) = 0.995 \) | \( P(\text{Positive} \mid \text{No Disease}) = 0.01 \) | \( 0.995 \times 0.01 = 0.00995 \) |

The total probability of a positive test result is the sum of the joint probabilities:
\[
P(\text{Positive Test}) = 0.00475 + 0.00995 = 0.0147
\]

### Now applying Bayes' Theorem:

\[
P(\text{Disease} \mid \text{Positive Test}) = \frac{0.00475}{0.0147} \approx 0.323
\]

### Final Answer:
The probability that a person has the disease given that the test result is positive is approximately **32.3%**.

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

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Here are five related questions you could explore:
1. How would the result change if the population had a higher prevalence of the disease?
2. What happens to the probability if the false positive rate increases?
3. Can you calculate the probability that a person does not have the disease given a negative test?
4. How does the effectiveness of a test influence the outcome of Bayes' Theorem?
5. What other real-world situations can Bayes' Theorem be applied to?

**Tip**: When dealing with probabilities, always identify the relevant prior probabilities and conditional probabilities before applying Bayes' Theorem.

A laboratory blood test is 95 percent effective in detecting a certain disease when it is, in fact, present. However, the test also yields a 'false positive' result for 1 percent of the healthy persons tested. If 0.5 percent of the population actually has the disease, what is the probability that a person has the disease given that the test result is positive?

Learn how to apply Bayes' Theorem to calculate the probability of having a disease given a positive test result. This problem covers conditional probability, false positives, and how to interpret medical test results using real-world data. Suitable for students in Grades 10-12.

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