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MHT CET · Maths · Probability

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 d2 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

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

Answer & Solution

Correct Answer

(A) \(\frac{1}{4}\)

Step-by-step Solution

Detailed explanation

\(P(D_2|T) = \frac{P(T|D_2)P(D_2)}{P(T|D_1)P(D_1) + P(T|D_2)P(D_2) + P(T|D_3)P(D_3)}\) \(P(D_2|T) = \frac{0.5 \times \frac{1}{3}}{0.7 \times \frac{1}{3} + 0.5 \times \frac{1}{3} + 0.8 \times \frac{1}{3}}\)