Question

A doctor assumes patient has $d_1, d_2,$ or $d_3$ with equal probability. A test is positive with probability 0.7 for $d_1$, 0.5 for $d_2$, and 0.8 for $d_3$. If the test is positive, what is the probability the patient has $d_2$?

• A. 1/4
• B. 1/2
• C. 1/5
• D. 1/4 Note: The options provided contain a duplicate '1/4'. Selecting (a) is appropriate.

Hint:

If the prior probabilities (base rates) are strictly identical for all cases, you can ignore them completely in the calculation! The posterior probability simply becomes the target likelihood divided by the sum of all likelihoods.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given prior probabilities of three diseases and the conditional probabilities of testing positive for each. We need to find the reverse conditional probability: given a positive test, what is the probability the patient has disease $d_2$? This strictly requires Bayes' Theorem.

Step 2: Key Formula or Approach:
Bayes' Theorem for three mutually exclusive events is:
$$P(d_2 | +) = \frac{P(+ | d_2) \cdot P(d_2)}{P(+ | d_1)P(d_1) + P(+ | d_2)P(d_2) + P(+ | d_3)P(d_3)}$$
(Where "$+$" represents the event that the test result is positive).

Step 3: Detailed Explanation:
1. Identify the prior probabilities:
Since the doctor assumes equal probability among three diseases:
$P(d_1) = P(d_2) = P(d_3) = 1/3$.
2. Identify the conditional probabilities (likelihoods):
$P(+ | d_1) = 0.7$
$P(+ | d_2) = 0.5$
$P(+ | d_3) = 0.8$
3. Calculate the total probability of a positive test (the denominator):
$$P(+) = (0.7 \times \frac{1}{3}) + (0.5 \times \frac{1}{3}) + (0.8 \times \frac{1}{3})$$
Since the $1/3$ factor is common to all terms, we can factor it out:
$$P(+) = \frac{1}{3}(0.7 + 0.5 + 0.8) = \frac{1}{3}(2.0)$$
4. Apply Bayes' Theorem:
$$P(d_2 | +) = \frac{0.5 \times \frac{1}{3}}{\frac{1}{3}(2.0)}$$
The common $1/3$ factor cancels out entirely from the numerator and denominator:
$$P(d_2 | +) = \frac{0.5}{2.0} = \frac{5}{20} = \frac{1}{4}$$

Step 4: Final Answer:
The probability is 1/4, matching option (a).