(C) Let $E_1$ be the event that the person suffers from the disease and $E_2$ be the event that the person does not suffer from the disease. Let $A$ b

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(C) Let $E_1$ be the event that the person suffers from the disease and $E_2$ be the event that the person does not suffer from the disease. Let $A$ be the event that the diagnostic test is positive. Given: $P(E_1) = 0.5 \% = 0.005$ $P(E_2) = 99.5 \% = 0.995$ $P(A|E_1) = 0.95$ $P(A|E_2) = 0.10$ We need to find $P(E_1|A)$. Using Bayes' Theorem: $P(E_1|A) = \frac{P(E_1) \times P(A|E_1)}{P(E_1) \times P(A|E_1) + P(E_2) \times P(A|E_2)}$ $P(E_1|A) = \frac{0.005 \times 0.95}{(0.005 \times 0.95) + (0.995 \times 0.10)}$ $P(E_1|A) = \frac{0.00475}{0.00475 + 0.0995}$ $P(E_1|A) = \frac{0.00475}{0.10425} = \frac{475}{10425} \approx 0.0455$

Class 12 Mathematics Probability Baye's theorem

Medium

$A$ diagnostic test has a probability of $0.95$ of giving a positive result when applied to a person suffering from a certain disease and a probability of $0.10$ of giving a positive result when given to a non-sufferer. It is estimated that $0.5 \%$ of the population are suffering from the disease. If this test is administered to a person from this population about whom there is no information relating to the incidence of this disease and the test gives a positive result,then the probability that the person is a sufferer is:

TS EAMCET 2020 All TS EAMCET

A
$0.9545$
B
$0.2194$
C
$0.0455$
D
$0.9499$

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Probability

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Baye's theorem

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MediumAP EAMCET 2024

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