SOLUTION: A certain virus infects one in every 500 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 pers

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Question 1150194: A certain virus infects one in every 500 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. Let A be the event "the person is infected" and B be the event "the person tests positive."
(a) Find the probability that a person has the virus given that they have tested positive.

(b) Find the probability that a person does not have the virus given that they have tested negative.

Answer by ikleyn(52776) About Me  (Show Source):
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A certain virus infects one in every 500 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.
Let A be the event "the person is infected" and B be the event "the person tests positive."
    (a)   Find the probability that a person has the virus given that he or she is tested positive.
    (b)   Find the probability that a person does not have the virus given that he or she is tested negative.

Solution

(a)  To make the solution easier, let assume that we have a population of 100,000 persons in the city.

      According to the input data,  1%2F500 of them, i.e.   100000%2F500 = 200 are infected.


      Of this 200 people, 90%, i.e.  0.9*200 = 180  are test positive.


      The problem asks in (a), what is the ratio of the number of those who are test positives AND have the virus 
      to the number of infected.


      It is obvious that this ratio is  P = 18%2F20 = 0.9 = 90%.    ANSWER


      In the compact form, a formula for the probability (a) is  P = %280.002%2A0.9%29%2F0.002 = 0.9.



(b)  The probability under question (b) is the ratio of the number of those who do not have a virus AND are tested negative
     to the number of those who do not have the virus.


     The number of those who do not have the virus is  100000 - 0.002*100000 = 100000 - 200 = 99800.
     
     Of them, the number of those who are tested negative is  (1-0.1)*99800 = 0.9*99800 = 89820.



     With these numbers, the probability (b) is

              P = 89820%2F99800 = 0.9.     ANSWER


     In the compact form, a formula for probability (b) is  P = %28%281-0.002%29%2A%281-0.1%29%29%2F%281-0.002%29 = 0.9.

Solved.