Quantcast
Student Doctor Network Forums - View Single Post - NBME #3 Biostat Question
View Single Post
Old 05-20-2012, 02:21 PM   #9
2K Member
 
mdeast's Avatar
 
Join Date: Jun 2009
Posts: 2,284
SDN 2+ Year Member
Default

Quote:
Originally Posted by Phloston View Post
The correct answer is B. The fact that the patient survived to one year is irrelevant, as would if he had survived two or three. Choice B represents the probability of surviving all four years.

This is not the same as asking the probability of landing three additional tails (1/8) if a previous one has already been landed. That's the logic that choice D would follow.

Choice B would be analogous to asking what's the probability of landing four tails in a row. This is independent of whether one, two or three have already been landed. The question's just asking how likely it is that this patient will live four years.
Don't think this reasoning is correct. Take the tails example. If the patient had died the first year, (i.e. heads), he'd have a 0% of making it to 4/4 years. If he lived (i.e. rolled tails), his probability of reaching 4 tails in a row would now rely solely on the last 3 spins (i.e. .5 x .5 x .5 = 1/8).

I actually looked this up online after doing the offline version of this morning. Most boards online say the answer is D and that the answer key is just incorrect. This happens a lot with bootlegged versions of these exams.
__________________
MS2 (almost MS3)
mdeast is offline   Reply With Quote