Question

1. a blood test indicates the presence of a particular disease (95\\%) of the time when the disease is actually present. the same test indicates the presence of the disease (0.5\\%) of the time when the disease is not actually present. one percent of the population actually has the disease.

calculate the probability that a person actually has the disease given that the test indicates the presence of the disease.

(a) 0.324
(b) 0.657
(c) 0.945
(d) 0.950
(e) 0.995

Explanation:

⚡ Using what you learned: Conditional Probability · 🆕 New: Bayes' Theorem
Updating beliefs with new evidence

Step 1: Define events and probabilities

Let \( D \) be the event that a person has the disease, and \( T \) be the event that the test is positive (indicates the presence of the disease).

From the problem statement:

• \( P(D) = 0.01 \) (one percent of the population has the disease)
• \( P(D^c) = 1 - 0.01 = 0.99 \) (the rest of the population does not have the disease)
• \( P(T \mid D) = 0.95 \) (true positive rate)
• \( P(T \mid D^c) = 0.005 \) (false positive rate of \( 0.5\% \))

We want to find the conditional probability \( P(D \mid T) \).

Step 2: Calculate the total probability of a positive test

To find the overall probability of testing positive, \( P(T) \), we combine the cases of testing positive with the disease and testing positive without the disease:

$$ P(T) = P(T \mid D)P(D) + P(T \mid D^c)P(D^c) $$

$$ P(T) = (0.95)(0.01) + (0.005)(0.99) $$

$$ P(T) = 0.0095 + 0.00495 = 0.01445 $$

Step 3: Apply Bayes' Theorem

Now, we calculate the probability that a person actually has the disease given a positive test result:

$$ P(D \mid T) = \frac{P(T \mid D)P(D)}{P(T)} $$

$$ P(D \mid T) = \frac{0.0095}{0.01445} \approx 0.6574 $$

Comparing this with the given options, \( 0.657 \) corresponds to option (B).

Answer:

(B) 0.657