Question 1170302: A medical test has been designed to detect the presence of a certain disease. Among those who
have the disease, the probability that the disease will be detected by the test is 0.96. However,
the probability that the test will erroneously indicate the presence of the disease in those who
do not actually have it is 0.03. It is estimated that 95% of the population who take this test do
not have the disease.
a. Construct a tree diagram for the above information.
b. If a test is administered to an individual, what is the probability that the test is positive?
c. If a test administered to an individual is negative, what is the probability that the person
actually has the disease?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Absolutely! Let's break down this problem step by step.
**a. Constructing the Tree Diagram**
* **Branches:**
* The first set of branches will represent whether a person has the disease or not.
* The second set of branches will represent the test result (positive or negative).
* **Notation:**
* D = Has the disease
* D' = Does not have the disease
* + = Test is positive
* - = Test is negative
* **Probabilities:**
* P(D') = 0.95 (95% do not have the disease)
* P(D) = 1 - P(D') = 1 - 0.95 = 0.05 (5% have the disease)
* P(+|D) = 0.96 (probability of positive test given the disease)
* P(-|D) = 1 - P(+|D) = 1 - 0.96 = 0.04 (probability of negative test given the disease)
* P(+|D') = 0.03 (probability of positive test given no disease)
* P(-|D') = 1 - P(+|D') = 1 - 0.03 = 0.97 (probability of negative test given no disease)
Here's the tree diagram:
```
/ + (0.96)
D (0.05)
/ \ - (0.04)
/
/
/
Start -------------------
\
\
\ / + (0.03)
D' (0.95)
\ - (0.97)
```
**b. Probability of a Positive Test**
To find the probability of a positive test, we need to consider both cases: having the disease and not having the disease.
* P(+) = P(D and +) + P(D' and +)
* P(+) = P(D) * P(+|D) + P(D') * P(+|D')
* P(+) = (0.05 * 0.96) + (0.95 * 0.03)
* P(+) = 0.048 + 0.0285
* P(+) = 0.0765
Therefore, the probability that the test is positive is 0.0765.
**c. Probability of Having the Disease Given a Negative Test**
We need to find P(D|-). We'll use Bayes' Theorem:
* P(D|-) = [P(-|D) * P(D)] / P(-)
First, let's find P(-):
* P(-) = P(D and -) + P(D' and -)
* P(-) = P(D) * P(-|D) + P(D') * P(-|D')
* P(-) = (0.05 * 0.04) + (0.95 * 0.97)
* P(-) = 0.002 + 0.9215
* P(-) = 0.9235
Now, let's apply Bayes' Theorem:
* P(D|-) = [(0.04) * (0.05)] / 0.9235
* P(D|-) = 0.002 / 0.9235
* P(D|-) ≈ 0.0021656
Therefore, the probability that a person has the disease given a negative test is approximately 0.0021656.
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