# Conditional Probability and Disease Testing

Suppose a diagnostic test for a particular disease has a $95$% chance of returning a true positive and a $10$% chance of a false positive. Suppose everyone in the population is screened for the disease and the true proportion of diseased individuals is $d$. All those who test positive on the first screening are tested again and of these return patients, $25$% test positive (assuming the two disease tests are independent). What is $d$?

My attempt at a solution:

First, we compute $\text{Pr}(\text{First Test Positive})$ in terms of $d$. We have by the law of total probability: $$\text{Pr}(\text{FTP}) = \text{Pr}(\text{FTP} | \text{Disease})\text{Pr}(\text{Disease}) + \text{Pr}(\text{FTP} | \text{No Disease})\text{Pr}(\text{No Disease})$$ This is given by: $$.95 d + .1(1-d) = .10 + .85d$$ Now, we know by the problem that: $$\text{Pr}(\text{STP} | \text{FTP}) = 0.25$$ and that these events are independent. So: $$\text{Pr}(\text{STP} | \text{FTP}) = \text{Pr}(\text{STP}) = 0.25$$ Moreover, it is clear that $\text{Pr}(\text{STP}) = \text{Pr}(\text{FTP})$ so: $$.10 + .85d = .25$$ and thus: $$d = .176$$

I am a bit concerned by the final three lines (of math). I am not sure whether it is the wording of the problem tripping me up or my lack of understanding of conditional probability.

• Now that I think of it, I believe my assumption that of the independence of the second test being positive given the person tested positive for the first step is kind of erroneous, i.e someone who tested positive for the first test should have a higher chance of having the disease and thus test positive for the second test? My intuition seems right but I understand also how unintuitive a subject probability is. – rubikscube09 Oct 3 '17 at 3:40
• Your intuition is okay, the problem was badly worded. – Graham Kemp Oct 3 '17 at 4:20

That is $\mathsf P(FTP, STP\mid D)=\mathsf P(FTP\mid D)\,\mathsf P(STP\mid D)\\\mathsf P(FTP, STP\mid D^\complement)=\mathsf P(FTP\mid D^\complement)\,\mathsf P(STP\mid D^\complement)$
Using $D$ for "Disease" and $D^\complement$ for "no disease" to save some space:
\begin{align}\mathsf P(STP\mid FTP) ~&=~\dfrac{\mathsf P(FTP, STP)}{\mathsf P(FTP)} \\[1ex]&=~ \dfrac{\mathsf P(STP\mid D)\mathsf P(FTP\mid D)\mathsf P(D)+\mathsf P(STP\mid D^\complement)\mathsf P(FTP\mid D^\complement)\mathsf P(D^\complement)}{\mathsf P(FTP\mid D)\mathsf P(D)+\mathsf P(FTP\mid D^\complement)\mathsf P(D^\complement)}\\[2ex] 0.25 ~&=~ \dfrac{0.95^2d+0.10^2(1-d)}{0.95d+0.10(1-d)}\end{align}