6
$\begingroup$

I have the following problem:

$0.5$% of a population are infected with a dangerous virus. A diagnostic test for the identification of the virus is positive in $99$% for infected people and in $2$% for not infected people.

Please estimate the probability that a person whose test was positive is infected with the virus.

And this is my solution.

Since $0.5$% of the population is infected, then $99.5$% is not infected. Since $99$% of the tests on infected people is positive, then $1$% is negative. Similarly, we can conclude that $98$% of the tests in not infected people is negative.

What we need to find, probability that a person whose test was positive is infected with the virus or in other words that a person is infected given that the test was positive, can be expressed (using the Bayes' theorem) as $$p( P_I \mid T_P) = \frac{p(P_I)\cdot p(T_P \mid P_I)}{p(T_P)}$$

Where $P_I$ means person is infected and $T_P$ means tests are positive.

Now, we know $p(P_I)$, that is $\frac{0.5}{100} = 0.005$, and we also know $p(T_P \mid P_I) = \frac{99}{100} = 0.99$. We only need to find $p(T_P)$, that is the probability that tests are positive. For this purpose, we can use the law of the total probability in the following way:

$$p(T_P) = p(P_I)\cdot p(T_P \mid P_I) + p(\overline{P_I}) \cdot p(T_P \mid \overline{P_I}) = 0.005 \cdot 0.99 + 0.995\cdot 0.02 = 0.02485$$

We can now plug the numbers in the first equation

$$p( P_I \mid T_P) = \frac{0.005 \cdot 0.99}{0.02485} = 0.19919517102615694$$

That is the probability that person is infected given that the tests are positive is roughly $20$%.

Is my solution correct?

This $20$% does not convince me honestly...

$\endgroup$
5
  • $\begingroup$ Since the disease is rare in the population, having a test that is 99% accurate has very low probability of giving a true positive when administered to one randomly picked person. Yes, your intuition is fooling you. The thing is that in real life the patient most often has more than one symptom (and hence a test is not given to a random subject). $\endgroup$
    – mmh
    Commented Oct 21, 2015 at 18:18
  • $\begingroup$ This counterintuitive result is one of the famous applications of Bayes' theorem. It has been recast into many similar forms, but the medical test form seems to be the most popular. Putting the modeling issues that @mmh has pointed out aside, the problem in this situation is that there are too many false positives. That is, although $98\%$ of the non-infected test negative, they constitute $99.5\%$ of the population. Thus about $2\%$ of the overall population are false positives while only about $2.5\%$ of the population test positive in the first place. $\endgroup$
    – Ian
    Commented Oct 21, 2015 at 18:22
  • $\begingroup$ Here is one way of looking at it. One in two hundred of the population are infected. After the test, one in five of those testing positive are infected. So the test increases the effectiveness of whatever is done next. There is a factor of $40$ improvement between the two ratios. (The false negatives could also be problematic by the way) $\endgroup$
    – Mark Bennet
    Commented Oct 21, 2015 at 18:28
  • $\begingroup$ Being a bit explicit about what I was saying before, if $T$ denotes positive test and $I$ denotes infection then $P(I \mid T)=\frac{P(T \mid I)P(I)}{P(T \mid I)P(I)+P(T \mid I^c)P(I^c)}=\frac{1}{1+\frac{P(T \mid I^c) P(I^c)}{P(T \mid I)P(I)}}$. As $P(I) \to 0$, this is approximately $\frac{1}{1+\frac{P(T \mid I^c)}{P(T \mid I)P(I)}}$. For this to be close to $1$, $\frac{P(T \mid I^c)}{P(T \mid I)}$ will need to be considerably less than $P(I)$. So the problem is really about the false positives, if $P(I)$ is sufficiently small. $\endgroup$
    – Ian
    Commented Oct 21, 2015 at 18:36
  • $\begingroup$ In fact we can even do the following. Let $\epsilon_i=P(I),\epsilon_p=P(T \mid I^c),\epsilon_n=P(T^c \mid I)$. These are small parameters for the overall infection rate, the false positive rate, and the false negative rate. Then $P(I \mid T)=\frac{1}{1+\frac{\epsilon_p (1-\epsilon_i)}{\epsilon_i (1-\epsilon_n)}}=\frac{1}{1+\frac{\epsilon_p}{\epsilon_i}}$ plus corrections on the order of $\epsilon_i \epsilon_p$ and $\epsilon_i \epsilon_n$. So when all three of these are small, the dominant effect is how $\epsilon_p$ compares to $\epsilon_i$. In this form I think the situation is pretty clear. $\endgroup$
    – Ian
    Commented Oct 21, 2015 at 18:57

1 Answer 1

Reset to default
4
$\begingroup$

Yes, about 20% is the correct answer.

One way to check this is to work out the expected fractions of the total population that are:

  • infected and test positive: 0.5% × 99% = 0.495%,
  • infected and test negative: 0.5% × 1% = 0.005%,
  • not infected and test positive: 99.5% × 2% = 1.98%, and
  • not infected and test negative: 99.5% × 98% = 97.52%.

Thus, the fraction of the total population that test positive is 0.495% + 1.98% = 2.475%. Yet clearly, out of that approx. 2.5%, only about one fifth (i.e. approx. 0.5% of the total population) are actually infected.

One trick that can sometimes help to make sense of problems like this is to convert the fractions to actual numbers of individuals. So let's assume that our total population consists of 20,000 people (which is just large enough to make all the fractions work out to a whole number of people). Then:

  • 100 out of these 20,000 people (0.5%) are infected.
    • 99 of these 100 infected people (99%) test positive.
    • 1 of these 100 infected people (1%) tests negative.
  • 19,900 out of these 20,000 people (99.5%) are not infected.
    • 398 of these 19,900 uninfected people (2%) test positive.
    • 19,502 of these 19,900 uninfected people (98%) test negative.

Thus, the total number of people who test positive is 99 + 398 = 497. Out of these 497 people, 99 are actually infected, while 398 are false positives.

Yet another way to quickly figure out the approximate result is to note that almost all people (99.5% ≈ 100%) are uninfected, and almost all of the infected test positive (98% ≈ 100%).

Thus, the fraction of false positives in the full population is approximately equal to the given fraction of false positives among the uninfected (2% × 99.5% ≈ 2%), while the fraction of true positives is approximately equal to the fraction of infected (99% × 0.5% ≈ 0.5%). Thus, the rate of false positives in the population (≈ 2%) is about four times the rate of true positives (≈ 0.5%), and so only about one fifth of all positive test results are true.

$\endgroup$
0

You must log in to answer this question.