My Math Forum Tough conditional probabilities problem

 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: 346 Thanks: 6 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: 346 Thanks: 6 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)

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