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Given the problem of a patient taking a test for a disease where having the disease is denoted by X and the a positive test is denoted by Y,

  • the rate of occurrence of the disease in the general population is 1%
  • The odds of getting tested positive is 90% if you have the disease
  • the probability of a false positive is 3%

is it appropriate to solve the following through rearranging the total probability

$P(A) = P(A|B)P(B) + P(A|Not B)P(Not B)$

Into

$P(A|B) = (P(A) -P(A|B)P(B))/P(Not B)$

and then getting P(B) from the total probability

$P(B) = P(B|A)P(A) + P(B| Not A)P(Not A)$

then given B is binary P(Not B) from

$1 = P(B) + P(Not B)$

and getting P(A|B) from Bayes

$P(A|B) = = P(A)P(B|A)/P(B)$

and then substituting that all back into the first equation to get the result

Any help on this would be much appreciated

Golden_Ratio
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WrongEngineer
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    This is hard to read, as you seem to switch notation from $X,Y$ to $A,B$ without explanation. In any case, this problem (with different numbers) has been asked over and over again on this site. Just do a simple search. here is one duplicate for example. – lulu Mar 11 '22 at 22:34
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    Perhaps you mean "probability" in place of "odds"? – Golden_Ratio Mar 11 '22 at 22:51
  • Refer to the probability tree at the top of, and scroll down to "false omission rate" in, this post. – ryang Mar 12 '22 at 03:25

1 Answers1

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We have by Bayes' rule

$$P(X|Y^c)=\frac{P(Y^c|X)P(X)}{P(Y^c|X)P(X)+P(Y^c|X^c)P(X^c)}.$$

Your setup has $P(X)=0.01,P(Y|X)=0.9,P(Y|X^c)=0.03.$

Golden_Ratio
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