SOLUTION: There are 100 patients in a hospital with a certain disease. Of these, 10 are selected to undergo a drug treatment that increases the percentage cured rate from 50 percent to 75 pe

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Question 1114805: There are 100 patients in a hospital with a certain disease. Of these, 10 are selected to undergo a drug treatment that increases the percentage cured rate from 50 percent to 75 percent. What is the probability that the patient received a drug treatment if the patient is known to be cured?
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
this is what i think it will be.

the probability that a patient receives the drug treatment is 10/100 = .1

the formula for that is p(D) = .1

10% of the time, the patient is cured using the drug, and 90% of the time the patient is cured without taking the drug.

the formula for that is p(C) = .1 * .75 + .9 * .5 = .525

the probability that a person is cured and has taken the drug is .1 * .75 = .075

the formula for that is p(C and D) = .075

given that a person is cured, what is the probability that the person took the drug.

the formula for that is p(D given C) = p(C and D) / p(C).

that becomes p(D given C) = .075 / .525 = .1428571429.

i usually try an experiment to see if what i did is reasonable.

my experiment for this problem is as follows.

assume 1000 patients.

10% of them receive the drug and 90% of them don't.

that makes 100 who received the drug and 900 who didn't.

75% of the ones who took the drug are cured = 75

50% of the ones who didn't take the drug are cured = 450.

total number of cured patients = 525.

525/1000 = .525, so p(C) looks like an accurate ratio.

given that a person is cured, what is the probability that the person took the drug.

525 were cured and, out of them, 75 took the drug.

75/525 = .1428571429, so p(D given C) looks good as well.

75 patients out of the 1000 gook the drug and were cured.

p(D and C) = 75 / 1000 = .075, so that ratio looks good as well.

the solution looks reasonable, so i would go with that.









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