Question 1200744
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P% of the country's population have symptoms of a certain disease. 
When tested, a person with    symptoms is diagnosed with a probability of p1, 
while        a person without symptoms tests positive with a probability of p2. 
(a) What is the probability that a randomly tested person will have a positive test result? 
(b) What is the probability that a person who tested positive actually has symptoms of the disease? 
([P, p1, p2] = [14, 0.96, 0.15])
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(a)  The logic of the solution is as follows:

         We have two disjoint sets: one "with the symptomps" and the other "without the symptoms".

         If a person belongs to the first set, he contributes to the sough probability 
         with the weight p1; 
         if a person belongs to the second set, he contributes to the sough probability 
         with the weight p2.


         It gives the <U>ANSWER</U> to question (a)  0.14*0.96 + (1-0.14)*0.15 = 0.2634.



(b)  In the second question, they want you calculate a <U>conditional probability</U>.

     So, this probability is the ratio of two values.
     The denominator is the probability that a randomly chosen person is tested positively.

     It is the probability which we just found out in part (a) above.


     The numerator is the measure of the intersection of two sub-sets: those who has symptoms AND, 
     at the same time, are tested positively. So, the probability in the numerator is

            0.14*0.96.


      Thus the final expression and the value for this conditional probability is

            {{{(0.14*0.96)/(0.14*0.96 + (1-0.14)*0.15)}}} = 0.51025   (rounded).    <U>ANSWER</U>
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Solved.