Biotech Values

[DewDiligence]

Re: Drug A or drug B?

Without loss of generality, say that 100 people in this patient pool take the test; within this group of 100, there will be 1 person who actually has the disease in question.

For the 1 person with the disease, the test will generate 0.1 false negatives and 0.9 true positives (a 10% false-negative rate).

For the 99 people without the disease, the test will generate 29.7 false positives and 69.3 true negatives (a 30% false-positive rate).

Thus, the test’s positive predictive value (in this patient pool) = true positives / total positives = 0.9/(0.9+29.7) = 0.9/30.6 = 2.94%. In other words, if a person in this patient pool tests positive, the probability that he actually has the disease is 2.94%.

(Although not necessary for solving your problem, the test’s negative predictive value in this patient pool = true negatives / total negatives = 69.3/(69.3+0.1) = 69.3/69.4 = 99.86%.)

If a person who tests positive in this patient pool takes drug A, he will be killed by side effects 20% of the time and will live the other 80% of the time for a survival rate of 80%.

If he takes drug B, he will live if he either: i) does not have disease in the first place (probability 97.06%); or ii) has disease and is cured (probability 2.94% x 70% = 2.06%). The sum of these cases is 99.12%, which is the survival rate for taking drug B.

Hence, drug B is clearly the better choice.