A test for a certain disease has a probability $\frac{4}{5}$ of detecting the disease when it ispresent and a probability of $\frac{1}{10}$ of falsely \detecting" it when it is not present.

The proportion of people afflicted by the disease for a given population is 15%.

If a person from the population is randomly selected and gives a positive result to the test, what is the probability that this person really has the disease? (Answer: 24/41).

I'm struggling to get that answer. I honestly just don't know how to approach this question.

  • 2
    $\begingroup$ Try applying these fractions/percentages to an actual population (say, 200 people). $\endgroup$ – Semiclassical Aug 16 '14 at 16:59

Let $D$ and $\neg D$ represent the events that the patient has and does not have the disease respectively, and let $T$ and $\neg T$ represent the event that the test is positive or negative respectively.

What do we want to compute?

$P(D \mid T)$

Unfortunately, all of our probabilities are in the form "test is (pos/neg) given patient (has/doesn't have) disease," which is "backwards" of what we want to compute. Bayes's Rule to the rescue!

\begin{align*}P(D \mid T) &= \frac{P(D \cap T)}{P(T)}\\ &= \frac{P(T \mid D) P(D)}{P(T)} \\&= \frac{P(T \mid D) P(D)}{P(T \mid D)P(D) + P(T \mid \neg D)P(\neg D)} \end{align*}


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