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 Concept:
We use Bayes' Theorem. Let $E$ be the event that the test is positive.

Step 2: Formula Application:
$P(d_1) = P(d_2) = P(d_3) = 1/3$. $P(E|d_1) = 0.7, P(E|d_2) = 0.5, P(E|d_3) = 0.8$. $P(d_2|E) = \frac{P(d_2)P(E|d_2)}{P(d_1)P(E|d_1) + P(d_2)P(E|d_2) + P(d_3)P(E|d_3)}$.

Step 3: Explanation:
$P(d_2|E) = \frac{(1/3)(0.5)}{(1/3)(0.7) + (1/3)(0.5) + (1/3)(0.8)}$ $P(d_2|E) = \frac{0.5}{0.7 + 0.5 + 0.8} = \frac{0.5}{2.0} = 1/4$.

Step 4: Final Answer:
The probability is 1/4.