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
- A \(\frac{1}{4}\)
- B \(\frac{1}{2}\)
- C \(\frac{1}{5}\)
- D \(\frac{1}{7}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
\(P(D_2|T) = \frac{P(T|D_2)P(D_2)}{P(T|D_1)P(D_1) + P(T|D_2)P(D_2) + P(T|D_3)P(D_3)}\) \(P(D_2|T) = \frac{0.5 \times \frac{1}{3}}{0.7 \times \frac{1}{3} + 0.5 \times \frac{1}{3} + 0.8 \times \frac{1}{3}}\)
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