A laboratory blood test is 95 % effective in detecting a certain disease when it is actually present. However, the test also yields a “false positive" result for 1 % of the healthy persons tested. If .5 % of the population actually has the disease, what is the probability a person has the disease given the test result is positive.

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### Probability Analysis of Disease Detection Using a Laboratory Blood Test

A laboratory blood test has a 95% effectiveness rate in detecting a certain disease when it is actually present. However, the test also yields a “false positive” result for 1% of healthy individuals tested. Given that 0.5% (or 0.005) of the population actually has the disease, we want to determine the probability that a person has the disease given that the test result is positive.

To solve this, we use Bayes' theorem, which incorporates conditional probabilities. Here is a breakdown of the relevant probabilities:

1. **Probability of a positive test given the disease is present (True Positive Rate):** 
   - \( P(\text{Positive} | \text{Disease}) = 0.95 \)

2. **Probability of a positive test given the disease is not present (False Positive Rate):**
   - \( P(\text{Positive} | \text{No Disease}) = 0.01 \)

3. **Probability of having the disease (Prevalence):**
   - \( P(\text{Disease}) = 0.005 \)

4. **Probability of not having the disease:**
   - \( P(\text{No Disease}) = 1 - P(\text{Disease}) = 1 - 0.005 = 0.995 \)

We are interested in finding \( P(\text{Disease} | \text{Positive}) \), the probability of having the disease given a positive test result.

According to Bayes' theorem:

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

Where \( P(\text{Positive}) \) is the total probability of a positive test result, which can be found using the law of total probability:

\[ 
P(\text{Positive}) = P(\text{Positive} | \text{Disease}) \cdot P(\text{Disease}) + P(\text{Positive} | \text{No Disease}) \cdot P(\text{No Disease}) 
\]

Plugging in the numbers:

\[ 
P(\text{Positive}) = (0.95 \cdot 0.005) + (0.01 \cdot 0.995)
Transcribed Image Text:### Probability Analysis of Disease Detection Using a Laboratory Blood Test A laboratory blood test has a 95% effectiveness rate in detecting a certain disease when it is actually present. However, the test also yields a “false positive” result for 1% of healthy individuals tested. Given that 0.5% (or 0.005) of the population actually has the disease, we want to determine the probability that a person has the disease given that the test result is positive. To solve this, we use Bayes' theorem, which incorporates conditional probabilities. Here is a breakdown of the relevant probabilities: 1. **Probability of a positive test given the disease is present (True Positive Rate):** - \( P(\text{Positive} | \text{Disease}) = 0.95 \) 2. **Probability of a positive test given the disease is not present (False Positive Rate):** - \( P(\text{Positive} | \text{No Disease}) = 0.01 \) 3. **Probability of having the disease (Prevalence):** - \( P(\text{Disease}) = 0.005 \) 4. **Probability of not having the disease:** - \( P(\text{No Disease}) = 1 - P(\text{Disease}) = 1 - 0.005 = 0.995 \) We are interested in finding \( P(\text{Disease} | \text{Positive}) \), the probability of having the disease given a positive test result. According to Bayes' theorem: \[ P(\text{Disease} | \text{Positive}) = \frac{P(\text{Positive} | \text{Disease}) \cdot P(\text{Disease})}{P(\text{Positive})} \] Where \( P(\text{Positive}) \) is the total probability of a positive test result, which can be found using the law of total probability: \[ P(\text{Positive}) = P(\text{Positive} | \text{Disease}) \cdot P(\text{Disease}) + P(\text{Positive} | \text{No Disease}) \cdot P(\text{No Disease}) \] Plugging in the numbers: \[ P(\text{Positive}) = (0.95 \cdot 0.005) + (0.01 \cdot 0.995)
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