0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – rubikscube09 Oct 3 '17 at 3:40
  • $\begingroup$ Your intuition is okay, the problem was badly worded. $\endgroup$ – Graham Kemp Oct 3 '17 at 4:20
1
$\begingroup$

You have been told to assume that the tests indepent, but indeed that is not a good assumption.   Rather, to solve the problem you need to assume that the tests are conditionally independent, given the same person.

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:

$$\begin{align}\mathsf P(STP\mid FTP) ~&=~\dfrac{\mathsf P(FTP, STP)}{\mathsf P(FTP)} \\[1ex]&=~ \dfrac{\mathsf P(STP\mid D)\mathsf P(FTP\mid D)\mathsf P(D)+\mathsf P(STP\mid D^\complement)\mathsf P(FTP\mid D^\complement)\mathsf P(D^\complement)}{\mathsf P(FTP\mid D)\mathsf P(D)+\mathsf P(FTP\mid D^\complement)\mathsf P(D^\complement)}\\[2ex] 0.25 ~&=~ \dfrac{0.95^2d+0.10^2(1-d)}{0.95d+0.10(1-d)}\end{align}$$

So just rearrange and solve.

$\endgroup$
  • $\begingroup$ Thank you. I did not realize that was what I needed to take into account. I find it odd that probability seems less intuitive than the functional analysis class I am currently taking. $\endgroup$ – rubikscube09 Oct 3 '17 at 5:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.