
Advanced Statistics Advanced Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
November 3rd, 2018, 03:33 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2  Tough conditional probabilities problem A patient comes into a doctor office exhibiting two symptoms: s1 and s2. The doctor has two possible diagnoses: disease d1 or disease d2. Assume that, given the symptoms, the patient must have either d1 or d2, but cannot have both. The following probabilities are given: P(s1d1) = 0.8 P(s1d2) = 0.4 P(s2d1) = 0.2 P(s2d2) = 0.6 P(d1) = 0.003 P(d2) = 0.007 Assume that s1 and s2 are conditionally independent, given the disease. 1) What are P(d1s1,s2) and P(d2s1,s2)? Answer: Code: P(d1s1,s2) = P(s1d1)P(s2d1)P(d1)/P(s1,s2) = P(s1d1)P(s2d1)P(d1)/(P(s1d1)P(s2d1)P(d1) + P(s1d2)P(s2d2)P(d2)) = (0.8)(0.2)(0.003)/((0.8)(0.2)(0.003)+(0.4)(0.6)(0.007)) = 0.22 P(d2s2,s2) = P(s1d2)P(s2d2)P(d2)/P(s1,s2) = P(s1d2)P(s2d2)P(d2)/(P(s1d1)P(s2d1)P(d1) + P(s1d2)P(s2d2)P(d2)) = (0.4)(0.6)(0.007)/((0.8)(0.2)(0.003)+(0.4)(0.2)(0.007)) = 0.78 = 1  P(d1s1,s2) P(cd1,t1) = 0.8 P(cd2,t1) = 0.1 P(cd1,t2) = 0.3 P(cd2,t2) = 0.6 Assume that event c is conditionally independent of the symptoms, given the disease and the treatment. What is the P(ct1,s1,s2)? I have been working on this problem for a week and I still couldn't solve question 2. Any tips would be much appreciated. Last edited by skipjack; November 3rd, 2018 at 05:23 PM. 
November 5th, 2018, 04:20 PM  #2 
Senior Member Joined: Aug 2008 From: Blacksburg VA USA Posts: 342 Thanks: 5 Math Focus: primes of course 
Part a) gave likelihood of which disease (d2 about 3 times more likely than d1) Part b) question focus is treatment 1. Knowing the probability of the actual disease, and how each responds to 1 of 2 specific treatments, provides all we need. So I think P(disease)*P(cure with that disease given a treatment) provides the answer. P(d1)*P(cd1,t1) + P(d2)*P(cd2,t1) = 0.22*0.80+ 0.78*0.10 = 0.254 So that's about a 1 in 4 chance of being cured. Of course, the patient has one of the 2 diseases. So their actual cure probability will either be 0.80 OR 0.10 respectively with t1 if they have d1 or d2. 
November 5th, 2018, 06:02 PM  #3 
Senior Member Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2 
I understand your logic, but I am unable to resolve the following: P(d1s1,s2)*P(cd1,t1) + P(d2s1,s2)*P(cd2,t1) = P(ct1,s1,s2) 
November 6th, 2018, 05:31 PM  #4 
Senior Member Joined: Aug 2008 From: Blacksburg VA USA Posts: 342 Thanks: 5 Math Focus: primes of course 
We know the treatment is t1, we know the cures by disease under that specific treatment, and the chance of each of the disease. P(xy,s1,s2) is equivalent to P(xy) as the stated condition is the patient presents with s1&s1, so it is a given restriction. Same for d2. Again, since it is stated that t1 is the treatment for the question, P(cd1,t1) becomes P(cd1) for that treatment. Same for d2. So we have P(d1)*P(cd1) + P(d2)*P(cd2) = P(c under stated conditions) = P(ct1,s1,s2) 

Tags 
conditional, conditionally, probabilities, problem, tough 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Problems with conditional probabilities  tvtokyo  Probability and Statistics  1  February 2nd, 2015 05:45 AM 
A question regarding conditional probabilities  lajka  Advanced Statistics  1  May 29th, 2011 02:24 PM 
Tough problem  Ashley  Calculus  1  December 3rd, 2009 06:47 PM 
A question regarding conditional probabilities  lajka  Algebra  1  December 31st, 1969 04:00 PM 