# NBME #3 Biostat Question

Discussion in 'Step I' started by thinkenergy, 05.19.12.

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1. ### thinkenergy 2+ Year Member

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This is from NBME #3:

Correct answer is B. But if patient survives the first year, wouldn't the first probability be 1? This is what I was thinking and I ended up choosing D, please help me understand why this is incorrect.

Thanks.

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3. ### ijnBanned 2+ Year Member

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Don't necessarily trust the keys out on the internet. It's D. For some reason Indian IMGs can't deal with basic probabilities well.

4. ### johndoe3344 5+ Year Member

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And yes, it's D.

5. ### PhlostonLifetime Donor 2+ Year Member

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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.

6. ### lord_jeebus和魂洋才 SDN Moderator 10+ Year Member

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I disagree, because this patient is already known to have survived year 1, the probability of this particular patient surviving year 1 is 100%, not the 80% probability for the population.

To make it about coin tosses:

If I have flipped tails once, what is the probability of flipping tails 4 times in a row total? The answer is (1/2)^3.

7. ### ijnBanned 2+ Year Member

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I guess I should revise my last comment to IMGs from India and Australia.

Kidding.

When you take NBME 3 online it never tells you which questions you got wrong and it never tells you which answer is correct. The only reason the "correct" answer is considered to be B by many IMGs is that there's an illegal/pirated NBME document that floats around their circles. That key has B as the answer. It's clearly incorrect, but because many students use those things to study they perpetuate the error.

NBME 4 has the exact same question where a patient has survived two years and then asks the probability of them living for 4 years. The answer in that question also was not the straight 4 year survival probability. You have to take into account that they lived through X many years.

8. ### loveoforganic-Account Deactivated- 2+ Year Member

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D for me as well

9. ### johndoe3344 5+ Year Member

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It's a conditional probability. P(4 year survival | 1 year survival) is not the same as P(4 year survival). Similarly, the probability of landing 4 tails in a row given 1, 2, or 3 has already landed is (1/2)^3, (1/2)^2, and (1/2)^1, respectively.

10. ### mdeast 5+ Year Member

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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.

11. ### PhlostonLifetime Donor 2+ Year Member

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Haha it's B, not D!

I gave my reasoning! And to whoever up above made the IMG comment, you're just embarrassing yourself!

12. ### VisionaryTicsSeñor Member 7+ Year Member

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Sorry, gotta disagree with you here, Phloston. It's clearly stating "Given that A is true (patient lived through year 1), what is the chance of B, C, and D happening (living to year 4)?"

13. ### ijnBanned 2+ Year Member

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If a person lives 4 years and 1 day post diagnosis, the chance they lived 4 years post diagnosis is 100%. Not 56.7%. You're being silly.

This EXACT same question appears in NBME 4 as well, except they've survived 2 years instead of 1. And in that instance the answer was multiplying the probability of surviving years 3 and 4. You assume a 100% survival rate for years 1 and 2, because they alive and stuff.

14. ### dr seuss 7+ Year Member

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Haha you're wrong, it's D!

Say for example there is a disease that kills 90% of people in the first year but once they make it past that only 10% of the survivors die each year. If you had a patient come to you after their first year and asked what their odds were of surviving year two, would you tell them they only have a 9% chance even though 90% of the other people in their situation live?

So if my hypothetical patient came to you at the end of year 3 and asked about surviving another year, you would tell them
7%? If I were that patient I would wonder why you only have a 7% success rate when all the other doctors had 90% of their patients survive year 3.

Last edited: 05.20.12
15. ### PhlostonLifetime Donor 2+ Year Member

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It's B, not D. Let me help you guys reason through it.

You have to understand that that data accounts for all patients, collectively, who received Tx for the cancer.

In the event of a single patient, however, his or her probability of survival remains unchanged from the beginning.

The perception of improved prognosis with respect to time is bound to a population-based function. The people who survive a given interval who account for the positive prognosis probabilities were going to survive anyway. Those who would die at year three, for instance, were going to die at year three regardless; the chance of survival of a latter patient is not augmented simply because he or she had lived to the second year. In terms of the population, probability of prognosis is ameliorated, but for a single individual, it is unchanged.

We could tell a patient that his or her chance of survival is gradually improved as time goes on, but in actuality, he or she, from day one, was going to live the same fixed length of time anyway.

This reminds me of:

From Wikipedia:

"The Monty Hall problem" because of its similarity to scenarios on the game show Let's Make a Deal, and its answer existed long before being posed to vos Savant, but was used in her column. Vos Savant answered arguing that the selection should be switched to door #2 because it has a 2/3 chance of success, while door #1 has just 1/3. Or to summarise, 2/3 of the time the opened door #3 will indicate the location of door with the car (the door you hadn't picked and the one not opened by the host). Only 1/3 of the time will the opened door #3 mislead you into changing from the winning door to a losing door. These probabilities assume you change your choice each time door #3 is opened, and that the host always opens a door with a goat. This response provoked letters of thousands of readers, nearly all arguing doors #1 and #2 each have an equal chance of success. A follow-up column reaffirming her position served only to intensify the debate and soon became a feature article on the front page of The New York Times. Among the ranks of dissenting arguments were hundreds of academics and mathematicians."

16. ### ijnBanned 2+ Year Member

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You don't understand conditional probabilities.

Here's a real life example with conditional probabilities and brain cancer survival.
TLDR: The fact you survived X years changes the probability that you will survive X+Y years.

17. ### PhlostonLifetime Donor 2+ Year Member

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A single patient who survives X+Y years was going to survive to X+Y years regardless. His or her probability of survival was unchanged from the beginning. In terms of the population as a whole, the conditional probabilities take into account those who have already died, so the latter are removed from the calculation. Therefore, of the remaining patients, they are of an increased prognosis pool, but on an individual level, nothing has changed.

Hope that helps

18. ### loveoforganic-Account Deactivated- 2+ Year Member

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Patients are not a homogenous sample. Individuals possess various characteristics that will make them more or less likely to survive for a given period. Lets say there are 5 black marbles and 5 white marbles. Black marbles have a structural defect that makes them prone to shatter into dust. White marbles do not. White marbles will therefore survive 10 minutes in a tumbler while black marbles will survive 5 minutes. You put them in a tumbler and pull one out. You have a 50/50 chance of getting either color marble. You run the tumbler for 6 minutes. The survival rate is 50% for a marble surviving to 6 minutes. The tumbler now has 100% white marbles. You run the tumbler for an additional 2 minutes. There is a 100% survival rate for a marble surviving from 6 minutes to 8 minutes. The new sample has different characteristics that will change the survival chances of the population of marbles

Edit: I should add that inherent characteristics of individuals aren't the only variability in disease course, obviously.

Last edited: 05.21.12
19. ### ijnBanned 2+ Year Member

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No. I really hope if you're dealing with cancer patients that you learn to understand this. You should not go tell your brain cancer patient who has survived 2 years that they only have a 27.6% chance of living until their 5th year. Their probability of living until their 5th year has increased to 76.2%. They'll appreciate hearing that.

20. ### PhlostonLifetime Donor 2+ Year Member

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Yes, it would be disheartening to inform them of that.

In terms of the new prognosis pool that the patient has become a part of, taking into account the removal of those who have already died, prognosis is improved. However, this patient was already part of the survival pool from the beginning. His or her chance of survival hasn't actually increased.

If we flip one tail and ask the probability of the next three flips being tails, the probability is 1/8. However, if we flip one tail and ask the probability of landing the fourth tails in a row, it's still 1/16.

I could see how this can be confusing for a lot of people.

21. ### loveoforganic-Account Deactivated- 2+ Year Member

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And tell me, how do you know which prognosis pool that patient is a part of, other than the one he is currently in?

Someone just send the bat signal for the psychology forum. They know stats better than anyone on these boards and can be the arbiter of justice

22. ### ijnBanned 2+ Year Member

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The question isn't asking for the patient's survival chance from year 0 to year 4. It's asking you for the patient's survival chance from year 1 to year 4. In your coin example, it'd be the probability of flipping 3 more tails if you've already flipped 1 tails.

It's a good rule on the STEP 1 exam to answer the question they have asked, not the question that you want to answer. I think you still have 8 more months to learn this though.

23. ### sanityonleaveAdrenaline Junkie 7+ Year Member

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Phloston--

You're definitely one of the smarter posters on the Step 1 forums, but you're wrong on this one mate. As someone else said, there's the exact same question on one of other the NBMEs, and you do not include the first year probability.

EDIT: Ha. I posted a link to the Monty Hall Problem as well without seeing that you had posted it above. It is true that some unequal distributions of probability are difficult to reason out, but this one is a pretty straightforward case of conditional probability.

EDIT2: Here's another way of looking at it -- let's say this isn't cancer but instead an infectious disease. Assume that 90% of patients die due to this disease and 10% of patients survive. Let's further assume that after being "cured" of this disease, a certain % of the population relapses each year and 100% of those people die.

The survival data would look like this (the numbers don't compute exactly, but you get the point):

10% 9% 8% 7%

If the patient survives the initial condition, their risk of dying per year is only whatever the relapse rate happens to be. Granted, their odds of survival decrease each year because they have to not relapse -- just like trying to flip tails 3 times in a row. I hope that example helped somewhat.

Last edited: 05.22.12
24. ### sanityonleaveAdrenaline Junkie 7+ Year Member

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This is incorrect. The events are independent. You could have flipped 500 tails in a row, and the probability of the 501st coming up tails is still 0.5.

Also, you shouldn't gamble -- the whole "law of averages" thinking leads to losing a lot of money gambling. I've been guilty of that one before .

25. ### mdquestion 5+ Year Member

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Pholston- To simplify this...let's rephrase the question to "If a patient survived 1 year and asked you (as the doctor) what is the probability that they will survive another year...what would you say?"

Would you say...

A. .80 x .875 = .70
B. .875

It's not A. The Patient has already survived 1 year. His chance of surviving to Year 1 is 100%. He becomes part of a new group of patients once reaching this point. As a doctor you would reply "Well, in a study, of the 400 people who survived 1 year, 50 people died during that year. Thus, your chances of dying this year are roughly 87.5". The question is actually asking for his chance of surviving Year 2. Not for his chance of surviving Years 1 and 2.

26. ### Suncrusher☣ ☣ ☣ ☣ ☣Lifetime Donor 5+ Year Member

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rofl @ the monty hall problem being posted by the same person whom it disproves

27. ### PhlostonLifetime Donor 2+ Year Member

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The idea/concept of conditional probabilities was not something that I had ever debated. I'm just pointing out that the NBME question doesn't say "which of the following is the probability that he lives the next three years?" It asks, "which of the following is the probability that he will survive 4 years?" And as I said before, if you flip one tails, the odds of landing three more are 1/8, but if you flip one tails, the odds of landing four are still 1/16. I understand what NBME is trying to ask, but the grammar doesn't support that.

28. ### PhlostonLifetime Donor 2+ Year Member

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She probably did receive quite a few responses like this guy's up above.

29. ### tjquinn 5+ Year Member

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You conveniently left out an important part of the question, it actually says: "based on this information, if a patient survives for 1 year, which of the following is the probability that he will survive for 4 years"

Also, as everyone else has been saying, if I flip one coin and get tails, the probability of getting tails again the next time is still 50%. The act of my previous coin flipping does not make more or less probable of getting heads or tails, the events are absolutely independent.

Now, if you were to say, what is the probability of flipping a coin 3 times in a row and getting tails each time, then it would be as you say, 1/8. However, if I then say, given that I have already flipped the coin 3 times, what is the probability that my 4th flip will be tails? This time I have specified my previous coin flippage, and thus any future coin flips are independent of what I just did; therefore, it will be a 50% chance of getting tails, given the information that I already flipped the coin 3 times and got tails.

In contrast, if I ask the question, what is the probability of flipping a coin 4 times in a row and getting tails each time, then the probability is, again as you say, 1/16. Again, this is much different if I say that I flipped a coin 3 times and got tails each time, what is the probability of getting tails on the next flip (i.e., the fourth flip), the answer is 50%.

The only time you can multiply probabilities the way you are suggesting is when you are predicting an outcome, not if something has already occurred. As people have pointed out, you are speaking exactly as gambling addicts speak; i.e., I bet my money on black and lost the last 10 times; therefore, I am more likely to win on black on the next round. This fallacy is due to forgetting that what happened in the past cannot influence any future events.

In contrast, if I am a gambler and I sit down at a table and I am trying to plan how I am going to lose my money, the accurate way of thinking would be as follows, "hmmm.. whats the probability that I will lose if I bet black 10 times in a row? ah, 1/20; therefore, it is highly unlikely to lose on black if I bet 10 times in a row!!!"

Now, the 50 year-old patient asks, "what is the probability I will survive to the age 54 doc?", well, then the answer is 0.8*0.875*0.9*0.9 = 0.567. This is because the patient has not begun to encroach on the first probability

If the patient comes back next year, and is still alive, and the he says, "You told me last time my probability to live to be 54 years was 0.567, I already lived one year, what is my probability of still living to 54 years old?" Then, you tell the patient, oh since you already lived that one year, your probability of living to 54 (i.e., 3 more years) is 0.875*0.9*0.9 = 0.7088

The KEY is that once an event has already occurred, it cannot influence the future event. The likelihood of flipping a tails 4 times in a row is 1/16, but the likelihood of flipping a tails 4 times in a row, knowing that I already received a tails on the first flip, is 1/8 since any future event is independent of the knowledge of the previous flip.

I feel like I'm rambling now

30. ### Suncrusher☣ ☣ ☣ ☣ ☣Lifetime Donor 5+ Year Member

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No, it asks "if he has already survived one year, what is the probability that he survives four years (i.e. three additional years)?" For all your focus on grammar, I think you're the only one in the thread who is interpreting the sentence wrongly and leaving out entire conditional clauses. Go back and read the exact wording again and then it will be clear why the pirated answer key has a different (wrong) answer from the real answer key.

Q: What are the odds that the patient will survive through year four if it is known that he is in the group that has already survived year one? A: 1.0*0.875*0.9*0.9 Q: What are the odds of landing three additional tails if you already landed one tails? A: 1.0*0.5*0.5*0.5.

Last edited: 05.22.12
31. ### PhlostonLifetime Donor 2+ Year Member

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We would all probably be better off by bringing out the friendship flag.

32. ### tjquinn 5+ Year Member

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Agreed, but your staunch resistance is inflammatory. It's not so much that you are wrong about this question, but that you are so confident that you will not acquiesce to basic fundamental probabilistic reasoning. A flag can and shall be raised, but some humility is required

33. ### PhlostonLifetime Donor 2+ Year Member

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There's no "staunch resistance." I've just concluded that we're going around in circles at this point. I don't think any of us had ever disagreed as to what this particular question may have been trying to ask, and although quite a few explanations/gestures have been offered up, I stand by my original reasoning based on the question's actual wording. It's not coincidental the fact that the answer key says B. Someone else saw it.

34. ### tjquinn 5+ Year Member

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Agreed, someone else did come to the same conclusion as you, but that doesn't make it the correct conclusion. Furthermore, the fact that this is such a commonly misunderstood concept, it is no wonder the most common mistake would be made on such a question. There are not a few explanations, there are multitudes of rephrasing one explanation vs. your single explanation. I'm not sure if you have read several of the posts above, including my own, but there is no question about what the question is "trying" to ask. The wording of the question is crystal clear:

"based on this information, if a patient survives for 1 year, which of the following is the probability that he will survive for 4 years"

You keep saying that the question says: "which of the following is the probability that he will survive 4 years?" Which it most certainly does NOT say. You are ignoring the MAIN part of the question stem, you cannot just read the part after the comma and conclude you have the right answer. This is what is so frustrating to, not only me, but other people trying to explain why your answer is in fact incorrect.

If you are answer "which of the following is the probability that he will survive 4 years?" then yes, you are absolutely correct. But, this is NOT the question being asked. Coming to a true conclusion about a true statement is not the same as coming to a true conclusion regarding the ACTUAL true question.

There is no circling, you are the only one that may be going in circles, this is a basic probability question that is taught in at least 8th grade, if not 7th grade.

35. ### johndoe3344 5+ Year Member

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You realize that on the actual ONLINE NBME form, people put D and got it correct, right?

You're basically saying that you're standing by the answer key of an OFFLINE version that some STUDENTS wrote up. Just so we're clear.

You also realize that every single other poster (without a single exception) is saying what we've all been saying in about 20 different ways? If that's not "staunch resistance", I don't know what is.

36. ### dr seuss 7+ Year Member

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Truth.

37. ### ImNotBritish.................... 7+ Year Member

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It's okay to be wrong. On the (admittedly few) occasions where I've seen you be mistaken and corrected on the forums, you always turn it into a "we'll just have to agree to disagree" type thing, even when it's pretty obvious who's in the right. And yet, you fight tooth and nail against misinformation in other threads. Seems contradictory...

38. ### openmindcnhealu

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Hey folks, Phloston is actually right. Based on the presented arguments, this is how the question writer intended to word the question:

"Based on this information, if the patient survives for 1 year, which of the following is the probability that he will survive for 3 additional years."

For the purpose of a board exam the question was ambiguously constructed. We process info clause by clause. The above rephrase has slight redundancy normal and essential to our lexicographic understanding. An important adjective was left out and a year added. This negated the presented condition creating ambiguity.

I won't be around to debate. ..... Phlostom = awesome!

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hi phloston2

40. ### Suncrusher☣ ☣ ☣ ☣ ☣Lifetime Donor 5+ Year Member

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not sure if amusing or pathetic...

41. ### thinkenergy 2+ Year Member

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I can't believe this thread is still going... LOL

42. ### illegallysmoothSmooth member 7+ Year Member

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Phloston, you are incorrect. I actually had this question today, and it is similar to a genetics question that I've gotten several times. Let's say a child's parents are heterozygous carriers of a recessive condition that is lethal in utero. The child is born healthy. What is the probability the child is a carrier? The grammar doesn't change anything. They gave you the conditions, you have to use them. If they tell you the child is healthy, you have to use that. If they tell you the man survived one year, you have to use that, you can't pretend it's one year ago and you don't know if he'll survive a year.

Following your logic, the child has a 50% chance of being a carrier. There's a 25% chance of homozygous recessive, and 25% chance of homozygous wild-type. The key flaw here is that there is no longer a 25% chance of being homozygous recessive because the child was born healthy. Now there is a 2/3 probability of being a heterozygous carrier and a 1/3 probability of being homozygous wild-type.

I think somewhere along this thread you realized you were wrong, so you flipped the script and starting arguing semantics.

Last edited: 05.25.12
43. ### tjquinn 5+ Year Member

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It's a lost cause... I too tried to explain in multiple ways.

44. ### loveoforganic-Account Deactivated- 2+ Year Member

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False dichotomy!

45. ### Suncrusher☣ ☣ ☣ ☣ ☣Lifetime Donor 5+ Year Member

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46. ### alternatego 2+ Year Member

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This is what great about SDN, I was looking for an answer to this question. BTW, I agree with ijn, I also think D is the answer.

47. ### thehundredthone 2+ Year Member

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Big useless bump perhaps, but I came across a similar question in NBME 6 (and maybe even 4/5). The probabilities are given for year 1-2, year 2-3, etc. So each is an individual probability and automatically assumes survival up to the previous year. E.g. the survival probability for year 3-4 means the person [i[has to have survived those 3 years[/i]. That's why the answer is to multiply the probability for each period of survival only as asked.