Question

A doctor assumes that patient has one of three diseases \(\mathrm{d} 1, \mathrm{~d} 2\) or d 3 . Before any test he assumes an equal probability for each disease. He carries out a test that will be positive with probability 0.7 if the patient has disease \(\mathrm{d} 1,0.5\) if the patient has disease d 2 and 0.8 if the patient has disease d3. Given that the outcome of the test was positive then probability that patient has disease d2 is

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{5}\)
• D. \(\frac{1}{7}\)

Hint:

Use Bayes theorem with equal prior probabilities for the three diseases.

Answer 1

The Correct Option is A

Step 1: Formula-first
$P(E_i\mid A)=\dfrac{P(E_i)P(A\mid E_i)}{\sum P(E_j)P(A\mid E_j)}$.

Step 2: Substitution
With equal priors this is the ratio of likelihoods: $0.5:(0.7+0.5+0.8)$.

Step 3: Simplify
$\dfrac{0.5}{2}=\dfrac14$, so option (A) is correct.

Final Answer:
With equal priors, the answer is the share of d2 in the total likelihood, 1/4. \[ \boxed{\text{(A) }\dfrac{1}{4}} \]