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I am struggling with an interview quiz question which starts with a standard conditional probability part:

To detect genetic defects, you are in charge of performing a test. You know that:

  • 1% of people have a genetic defect

  • 99.9% of tests for the gene detect the defect (true positives)

  • 5% of the tests give a positive result even though there is no defect (false positives)

Given that the condition of a patient is known, the results of multiple tests are independent

a) If a person gets a positive test result, what is the probability that he/she actually has the genetic defect?

For a), I would argue that this is standard conditional probability and can be solved with Bayes Rule. Let's call the event "$+$" when a person tests positive, "$d$" when a person has the disease and "$\bar{d}$" when a person does not have the disease.

Now we are looking for $P(d\mid+)$, so the conditional probability that someone actually has the disease when he tests positive. Given Bayes, that is

\begin{align} P(d\mid+) & = \frac{P(+\mid d)}{P(+)}\cdot P(d) = \frac{0.999}{P(+\mid d)\cdot P(d) + P(+\mid\bar{d})\cdot P(\bar{d})}\cdot 0.01 \\[10pt] & = \frac{0.999}{0.999\cdot 0.01 + 0.05\cdot 0.99}\cdot 0.01 \approx 0.168. \end{align}

So the probability is roughly 17%.

What is more complicated for me is b):

b) If a person gets a positive result in his/her first test, what is the probability of having a positive result in his/her second test?

I would argue that we are looking for $P(++|+)$, i.e. the probability that someone tests positive the second time under the condition that he tested positive the first time.

So we can apply Bayes again and get $\frac{P(+|++)}{P(+)}\cdot P(++)$. I'd argue that $P(+|++)$ is always 1 and that $P(++) = P(+)\cdot P(+)$. We can cancel out one $P(+)$ and this leaves us with $P(++|+) = P(+)$.

On the one hand this looks reasonable given the independence proclaimed, on the other hand it feels counter-intuitive that the probability would only be 6% (on the third hand, we are talking about statistics here and that never went well together with common sense for me :-)).

Thoughts?

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Denote events $D$, $\bar D$ as the patient having the defect and not having the defect, respectively. Let $P_i$ denote the event that test $i$ is positive, and $\bar P_i$ the event that test $i$ is negative.

Then the desired probability is $$\Pr[P_2 \mid P_1] = \frac{\Pr[P_2 \cap P_1]}{\Pr[P_1]} = \frac{\Pr[P_2 \cap P_1 \mid D]\Pr[D] + \Pr[P_2 \cap P_1 \mid \bar D]\Pr[\bar D]}{\Pr[P_1 \mid D]\Pr[D] + \Pr[P_1 \mid \bar D]\Pr[\bar D]}.$$ Since $P_i$ are conditionally independent given the defect status, we have $$\Pr[P_2 \cap P_1 \mid D] = (\Pr[P_i \mid D])^2.$$ Then, given $$\Pr[D] = 0.01, \quad \Pr[P_i \mid D] = 0.999, \quad \Pr[P_i \mid \bar D] = 0.05,$$ we easily obtain $$\Pr[P_2 \mid P_1] = \frac{(0.999)^2(0.01) + (0.05)^2(1-0.01)}{(0.999)(0.01) + (0.05)(1-0.01)} \approx 0.209363.$$ This number is small because the prevalence of defects is so rare, and the false positive rate is much higher than the prevalence. Therefore, a positive result is more likely to result from a false positive, and a second test is not terribly likely to come back positive.

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I am guessing that whoever wrote the question intended the tests to be conditionally independent given that the person is well and also given that the person is ill. They ought to have said that.

If we assume conditional independence, then, given that the person is well, the probability of two independent positive tests is just the square of the probability of a positive test, given that the patient is well. And the probability of two independent positive tests, given that the patient is ill, is the square of the probability of a positive test given that the patient is ill.

So just use Bayes's rule as you did originally, but with those squares of conditional probabilities.

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  • $\begingroup$ If a plus after the solidus means the test result is positive why would a plus before the solidus means the patient is actually ill? $\endgroup$ – A. Smith Jan 26 '17 at 19:19
  • $\begingroup$ @A.Smith : Because in the present context, no other meaning makes sense. $\endgroup$ – Michael Hardy Jan 26 '17 at 23:14

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