Replies to post #128850 on Biotech Values

[TastyTheElf]

the probability that he actually has the disease is 2.94%.

You're right. I forgot to add the 0.9% to the denominator.

[iwfal]

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.

Correct of course - and stated more dramatically:

Calculated the correct way the odds of death are more than 20x greater for Drug A than they are for Drug B.

But the human intuitional method can incorrectly get a very very different comparison of the risks of Drug A vs Drug B(as the good sport Tas showed - e.g. about equal odds for drug a and drug b).