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.
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.
A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
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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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F45fdd75f-924d-4ac4-9d15-03fc1f26bc02%2F89d37f9b-69b3-4963-bc3c-94de765c005d%2Fayzold6_processed.png&w=3840&q=75)
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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