Question

a certain virus infects one in every 600 people. a test used to detect the virus in a person is positive 90% of the time if the person has the virus and 10% of the time if the person does not have the virus. (a) find the probability that a person has the virus given that they have tested positive. $.0148026316$correct (b) find the probability that a person does not have the virus given that they have tested negative.

Answer 1

1. the probability that a person has the virus given that they have tested positive is  0.0151.
2. the probability that a person does not have the virus given that they have tested negative is  0.9999

P(A) = 1/600 = 0.0017

P(B) = 0.9 * 0.0017 + 0.1 * (1 - 0.0017) = 0.1014

A) P (has the virus | tested positive) = P (tested positive | has the virus) ×  

                                                          P (has the virus)/ P (tested positive)

                                                         = 0.9 × 0.0017/0.1014  

                                                         = 0.0151

B) P (does not have the virus | tested negative) = P (tested negative | does not have the virus) × P (does not have the virus)/ P (tested negative)

= (1 - 0.1) *× (1 - 0.0017)/ (1 - 0.1014)  

= 0.9999

Probability is the department of mathematics regarding numerical descriptions of ways likely an occasion is to occur, or how possibly it's far that a proposition is genuine. The possibility of an occasion varies between zero and 1, wherein, roughly speaking, 0 suggests the impossibility of the occasion and 1 shows certainty. The better the possibility of an event, the more likely it is that the event will arise.  

A simple instance is the tossing of an honest (unbiased) coin. since the coin is truthful, the 2 results ("heads" and "tails") are both equally likely; the possibility of "heads" equals the chance of "tails"; and considering the fact that no different results are feasible, the possibility of both "heads" or "tails" is 1/2 (that could additionally be written as 0.5 or 50%).

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