# Conditional probability, possibly related to Bayes' Theorem

Doctors test for tuberculosis using two tests: an inaccurate skin PPD test, and a chest X-ray. The chance that the PPD test gives a healthy person a positive result is 5%. The chance a sick person tests negative is 1%.

Assume the chance that a person with a positive PPD result has TB is 5%. What is the tuberculosis rate in the general population?

I don't know how to interpret the part "assume the chance that a person with a positive PPD result has TB is 5%" does this mean P(PPD tested positive AND Being TB positive) = P(PPD+ and TB+) or P(PPD+|TB+) or Probability of being TB positive GIVEN that PPD tested positive = P(TB+|PPD+)?

I'm more inclined to interpret the sentence as getting positive TB Given that PPD tested positive: P(TB+|PPD+)=0.05 therefore by using Bayes' Theorem I got the set up to be: $\frac {P(PPD+|TB+)P(TB+)}{P(PPD+)}=0.05$

Solving for P(TB), here is where I'm stuck as I don't know how to get P(PPD+|TB+). For P(PPD+) I got 0.05*0.95+0.01*0.99. So I tried a little algebraic manipulation:

P(PPD+|TB+)P(TB+)=0.05*P(PPD+)

P(TB+)= 0.00287*P(PPD+|TB+)

is my reasoning correct?

• It might help if you would tell us what events your abbreviations stand for instead of making us guess. I';m guessing that PPD+ means that the person has a positive PPD test result — is that right? I can't guess what events TB and TB+ denote. I guess one of them is "the person has tuberculosis" but which one? And what does the other one mean?
– bof
Commented Jul 7, 2018 at 6:57
• Positive PPD includes both 'Pos PPD & TB' and 'Pos PPD & no TB'. Commented Jul 7, 2018 at 6:58
• @bof my apology, I mistyped TB+. Commented Jul 7, 2018 at 7:40
• The problem gives $P(\text{PPD$-$} \mid \text{TB+}) = 0.01$, so $P(\text{PPD+} \mid \text{TB+}) = 1-0.01 = 0.99$. Commented Jul 7, 2018 at 8:05
• Please choose a more descriptive title for your question. Commented Jul 7, 2018 at 8:16

Interesting: This problem is sort of backwards from the usual ones, where you're given the specificity and sensitivity of the test, and the true prevalence of the disease, and you're expected to obtain the probability that a positive result is a true positive.

Here, instead, you're given the specificity ($95$ percent) and sensitivity ($99$ percent) as usual, but you're given the probability that a positive result is a true positive, and you're asked to obtain the true prevalence of the disease.

Nonetheless, they can be worked in more or less the same basic ways. One avenue is to imagine a generic population, where $q$ is the proportion of the population that has TB; that's what we're going to solve for. From that portion, $0.99q$ tests positive. From the $1-q$ that doesn't have TB, $0.05(1-q)$ also tests positive. That makes $0.99q + 0.05(1-q) = 0.05+0.94q$ that tests positive overall. Since we're told that $5$ percent of those that test positive actually have TB, we must have

$$0.99q = 0.05(0.05+0.94q)$$

from which we can solve for $q$.

Another possibility is to use Bayes's Formula directly:

$$0.05 = P(\text{has TB} \mid \text{tests positive}) = \frac{P(\text{tests positive} \mid \text{has TB}) P(\text{has TB})} {P(\text{tests positive})}$$

where

\begin{align} P(\text{tests positive}) & = P(\text{tests positive} \mid \text{has TB})P(\text{has TB}) \\ & + P(\text{tests positive} \mid \text{doesn't have TB}) P(\text{doesn't have TB}) \end{align}

Filling in the values we have gives us

$$0.05 = \frac{0.99q}{0.99q + 0.05(1-q)}$$

and again we can solve for $q$. It's essentially the same equation, but we arrived at it more heuristically in one case, and more symbolically in the other.

• WOW! Such a beautiful solution, thank you so much! I probably spent 2 hours trying setup some sort of equation to solve for the proportion but this is perfect! Commented Jul 7, 2018 at 8:36