Question 1191926
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A certain disease has an incidence rate of 0.5%. 
The false-negative rate on a test for the disease is 7%; 
the false positive rate is 5%. 
Compute the probability that a person who tests positive actually has the disease. 
(You may find it useful to construct a probability contingency table.)
The probability a person who tests positive actually has the disease is
Give your answer accurate to at least 3 decimal places
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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 0.005*X;  the number of those who has no the disease is 0.995*X.


     The number of those who has    the disease and tests positive is  0.005*(1-0.07)X = 0.005*0.93*X.

         To calculate this amount, I excluded from 0.005X, the number of sick persons, the value 0.005X*0.07, 
         which represents the part of sick who are tested as healthy (i.e. are false-negative).


     The number of those who has no the disease and (or "but") tests positive is  0.995*0.05*X              (false test positive).


So the number of those who tests positive is the sum  0.005*0.93*X + 0.995*0.05*X.


They want you calculate  the conditional probability


    P = {{{tests_positive_AND_has_the_disease/tests_positive}}} = {{{(0.005*0.93*X)/(0.005*0.93*X + 0.995*0.05*X)}}} = cancel X in the numerator and in the denominator = 

                   = {{{(0.005*0.93)/(0.005*0.93 + 0.995*0.05)}}} = {{{0.00465/(0.00465+0.04975)}}} = use your calculator = 0.085478  (rounded).    <U>ANSWER</U>


<U>ANSWER</U>.  This probability is  0.085478.
</pre>

Solved.