Question 1191803
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A certain disease has an incidence rate of 0.9%. 
If the false negative rate is 4% and the false positive rate is 5%, 
compute the probability that a person who tests positive actually has the disease.
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My previous solution was incorrect,  as tutor @math_tutor2020  noticed it.


Therefore,  I removed / (deleted / erased)  that version and placed here a new one,  corrected and fixed. 


So now you see the corrected and fixed version.    Thanks to tutor  @math_tutor2020.



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            This problem is about calculating the  CONDITIONAL  PROBABILITY.


            They want you calculate this ratio    {{{tests_positive_AND_has_the_disease/tests_positive}}}.



<pre>
Let X be the total population.


Then the number of those who has the disease is 0.009*X;  the number of those who has no the disease is 0.991*X.


     The number of those who has    the disease and tests positive is  0.009*(1-0.04)X = 0.009*0.96*X  (true  test positive)

                (to get this amount, I exclude from the sick persons, 0.009*X, the number of false negative 0.009*x*0.04.


     The number of those who has no the disease and tests positive is  0.991*0.05*X                    (false test positive).



So the number of those who tests positive is the sum  0.009*0.96*X + 0.991*0.05*X.


They want you calculate  the conditional probability


    P = {{{tests_positive_AND_has_the_disease/tests_positive}}} = {{{(0.009*0.96*X)/(0.009*0.96*X + 0.991*0.05*X)}}} = cancel X in the numerator and in the denominator = 

                   = {{{(0.009*0.96)/(0.009*0.96 + 0.991*0.05)}}} = {{{0.00864/(0.00864+0.04955)}}} = use your calculator = 0.1485  (rounded).    <U>ANSWER</U>


<U>ANSWER</U>.  This probability is  0.1485.
</pre>

Solved.