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Old 10-15-2010
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Stats Conditional Probability after surviving the first year?

hi guys !!

can anybody explain me how to do questions like these...

If a person has the probability of surviving 0-1 year is 80 percent, 1-2 year 87.5%, 2-3 year 90%, 3-4 year 90% what is the probability of him surviving 4 years if the patient survives for 1 year???
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Old 10-15-2010
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There are 2 probability rules:
  • For independent events (across multiple unrelated events): MULTIPLY
  • For mutually exclusive events (within a single event): ADD
This seems like a mutually exclusive event. So, we'll add the probability of surviving for one year (0.8) to the probability of surviving for 4 years (0.9). Total is 0.17 which is 170%.

Not sure about this though.

Another way that crossed my mind is to add the 3 probability chances after 1 year and get the average. (87.5 + 90 + 90) / 3 = 89.17%.

I'm not too good at those type of questions, so excuse me if both methods are incorrect!!!
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Old 10-15-2010
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Quote:
Originally Posted by usmleguy View Post
hi guys !!

can anybody explain me how to do questions like these...

If a person has the probability of surviving 0-1 year is 80percent, 1-2 year 87.5%, 2-3 year 90%, 3-4 year 90% what is the probability of him surviving 4 years if the patient survives for 1 year???
thanks for asking this Q...m also too weak with this kinda stuff.....
hope someone will nicely n easily explain this..
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Old 10-15-2010
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Embarrassed Is this valid?

The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. The notation for conditional probability is P(B|A), read as the probability of B given A.

so Probability of him surviving till year 2 is conditional on the fact that he has all ready survived year 1 and probability of that happening will be: 87.5/80 and we can do the same and then use multiplication rule and thats it. But this question i think is flawed as by common sense the probability should be decreasing not increasing as is the case here so my formule is not working! It will work if it was 80/87.5.

But thats just my opinion. I am so eager to hear from someone who really knows this stuff.
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Old 10-15-2010
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Don't make this question more complicated. We tend to get bogged down by questions when we think too much. Look, they told you that he survived the first year and they want to know what's the chance he'll survive 2nd, 3rd and 4th years. I know mutually exclusive means one or the other which normally in life and death situations would make sense, but in this case surviving the 4th year would depend on surviving the 2nd and 3rd years. This means these are dependent events. Dependent would mean nonindependent so what do you do for nonindependent events. Multiply! It says multiply for nonindependent events in the probability section of Kaplan notes. Just multiply the last three (0.875 * 0.90 * 0.90 = 0.70875) so appx 71%. I've gotten stars on my last two NBME's for Behavioral Science so I'm assuming this is right.
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3 candidates for president, labeled A, B,C lie 5%, 10%, 15% of the time respectively, tell half truths 40%, 45%, 50% of the time respectively and tell the truth 55%, 45%, and 35% respectively. Assume A tells twice as many statements as B and 3 times as many as C.

If a lie was told by a candidate, what is the probability it was from A? What percentage of all statements are half-truths?

OK..
So I proceeded with using the conditional probability formula: P(A|B) = P(AB)/P(B) but expanded it using the idea of Bayes Theorem:

P(A|E) = P(A)P(E|A)/P(A)P(E|A)+P(B)P(E|B)+P(C)P(E|C)
where A is the one we want and E is some subset.

So I got (1/3)(5/100)/(1/3)(5/100)+(1/3)(10/100)+(1/3)(15/100)

This will equal .1666666 but I times it by two since A issues twice as many statements as B. So the answer is 1/3. But I'm not sure if this is correct interpretation.

The second part of the question simply confuses me, I don't know where to proceed with that one.
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