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?
0.47
0.49
0.51
0.58
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:
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:
Where P(A') = 1 - P(A) = 0.90 (90% of the people do not have the disease)
Substituting the values, we get:
Now, applying Bayes' theorem:
Therefore, the probability that the person actually has the disease given a positive test result is 0.47.
Study 40% syllabus and score up to 100% marks in JEE
5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.820C. If Na2SO4 is 81.5% ionised, the value of x (K
A capacitor is made of two square plates each of side 'a' making a very small angle
A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO3 to give fraction A. The leftover organic phase was extracted with d