Question

A study concluded that among people with a certain​ virus, 99.2​% of tests conducted were​ (correctly) positive, while for people without the​ virus, 97.9​% of the tests were​ (correctly) negative. If 32​% of patients actually carry the​ virus, what's the probability that a patient testing negative is truly free of the​ virus?

Answer 1

Answer:

Following are the solution to the question:

Step-by-step explanation:

The probability of virus  P(V) =0.32  

No virus  P(NV) =0.66

Patient testing of negative P(TN) = virus, and negative testing+no virus and negative testing

$=0.32(1-0.992)+0.66 \times 0.979 \\\\=0.00256+0.64614\\\\=0.6487$

This is why the likelihood of a patient being free of the virus is negative.

$= \frac{P(NV) \times\ tested \ negative }{P(TN)}$  

$= \frac{0.66 \times 0.979}{0.6487}\\\\= \frac{0.64614}{0.6487} \\\\=0.996053646$