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 probability of a false positive is 3%, - the odds of getting tested positive is 90% if you have the disease

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 probability of a false positive is 3%
- the odds of getting tested positive is 90% if you have the disease
is it appropriate to solve the following through rearranging the total probability
$P\left(A\right)=P\left(A|B\right)P\left(B\right)+P\left(A|NotB\right)P\left(NotB\right)$
Into
$P\left(A|B\right)=\left(P\left(A\right)-P\left(A|B\right)P\left(B\right)\right)/P\left(NotB\right)$
and then getting P(B) from the total probability
$P\left(B\right)=P\left(B|A\right)P\left(A\right)+P\left(B|NotA\right)P\left(NotA\right)$
then given B is binary P(Not B) from
$1=P\left(B\right)+P\left(NotB\right)$
and getting P(A|B) from Bayes
$P\left(A|B\right)==P\left(A\right)P\left(B|A\right)/P\left(B\right)$
and then substituting that all back into the first equation to get the result
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Quinn Alvarez
We have by Bayes' rule
$P\left(X|{Y}^{c}\right)=\frac{P\left({Y}^{c}|X\right)P\left(X\right)}{P\left({Y}^{c}|X\right)P\left(X\right)+P\left({Y}^{c}|{X}^{c}\right)P\left({X}^{c}\right)}.$
Your setup has $P\left(X\right)=0.01,P\left(Y|X\right)=0.9,P\left(Y|{X}^{c}\right)=0.03.$