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 November 3rd, 2018, 02:33 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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(s1|d1) = 0.8 P(s1|d2) = 0.4 P(s2|d1) = 0.2 P(s2|d2) = 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(d1|s1,s2) and P(d2|s1,s2)? Answer: Code: P(d1|s1,s2) = P(s1|d1)P(s2|d1)P(d1)/P(s1,s2) = P(s1|d1)P(s2|d1)P(d1)/(P(s1|d1)P(s2|d1)P(d1) + P(s1|d2)P(s2|d2)P(d2)) = (0.8)(0.2)(0.003)/((0.8)(0.2)(0.003)+(0.4)(0.6)(0.007)) = 0.22 P(d2|s2,s2) = P(s1|d2)P(s2|d2)P(d2)/P(s1,s2) = P(s1|d2)P(s2|d2)P(d2)/(P(s1|d1)P(s2|d1)P(d1) + P(s1|d2)P(s2|d2)P(d2)) = (0.4)(0.6)(0.007)/((0.8)(0.2)(0.003)+(0.4)(0.2)(0.007)) = 0.78 = 1 - P(d1|s1,s2) 2) The doctor has the choice of two treatments t1 and t2. (It is not an option to do both.) Let c be the event that the patient is cured. The following probabilities are given: P(c|d1,t1) = 0.8 P(c|d2,t1) = 0.1 P(c|d1,t2) = 0.3 P(c|d2,t2) = 0.6 Assume that event c is conditionally independent of the symptoms, given the disease and the treatment. What is the P(c|t1,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 04:23 PM. November 5th, 2018, 03:20 PM #2 Senior Member   Joined: Aug 2008 From: Blacksburg VA USA Posts: 354 Thanks: 7 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(c|d1,t1) + P(d2)*P(c|d2,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. Thanks from zollen November 5th, 2018, 05:02 PM #3 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 I understand your logic, but I am unable to resolve the following: P(d1|s1,s2)*P(c|d1,t1) + P(d2|s1,s2)*P(c|d2,t1) = P(c|t1,s1,s2) November 6th, 2018, 04:31 PM #4 Senior Member   Joined: Aug 2008 From: Blacksburg VA USA Posts: 354 Thanks: 7 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(x|y,s1,s2) is equivalent to P(x|y) 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(c|d1,t1) becomes P(c|d1) for that treatment. Same for d2. So we have P(d1)*P(c|d1) + P(d2)*P(c|d2) = P(c| under stated conditions) = P(c|t1,s1,s2) Tags conditional, conditionally, probabilities, problem, tough Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post tvtokyo Probability and Statistics 1 February 2nd, 2015 04:45 AM lajka Advanced Statistics 1 May 29th, 2011 01:24 PM Ashley Calculus 1 December 3rd, 2009 05:47 PM lajka Algebra 1 December 31st, 1969 04:00 PM

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