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In a city, 10% of the population has a particular disease. A diagnostic test is available for the disease which is correct 80% of the time for people who have the disease and 90% of the time for people who do not have the disease. A person is selected randomly from the city and tested for the disease. If the test result is positive, what is the probability that the person actually has the disease? 

Option: 1

0.47


Option: 2

0.49


Option: 3

0.51


Option: 4

0.58


Answers (1)

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Let A be the event that a person has the disease, and B be the event that the test result is positive. 

We need to find P(A|B), the probability that the person actually has the disease given that the test result is positive. 

Using Bayes' theorem, we have:

P(A / B)=\frac{P(B / A) \times P(A)}{P(B)}

Where P(B|A) is the probability that the test result is positive given that the person has the disease, P(A) is the prior probability of a person having the disease, and P(B) is the probability of a positive test result. 

From the problem statement, we have: 

P(A) = 0.1 

P(B|A) = 0.8 (the test is correct 80% of the time for people who have the disease) 

P(B|A') = 0.1 (the test is incorrect 10% of the time for people who do not have the disease) We can calculate P(B) using the law of total probability:

P(B)=P(B / A) \times P(A)+P\left(B / A^{\prime}\right) \times P\left(A^{\prime}\right)

Where P(A') = 1 - P(A) = 0.90 (90% of the people do not have the disease) 

Substituting the values, we get:

\\{P(B)=0.8\times 0.1+\ 0.1\times 0.9}\\ \\\Rightarrow {P(B)=0.17}

Now, applying Bayes' theorem:

\begin{aligned} & P(A / B)=\frac{0.8 \times 0.1}{0.17} \\ \\& \Rightarrow P(A / B)=0.47 \end{aligned}

Therefore, the probability that the person actually has the disease given a positive test result is 0.47.

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seema garhwal

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