A doctor knows from experience that the probability a patient shows side effects

Question

A doctor knows from experience that the probability a patient shows side effects on a particular drug is 0.15. As- sume that probability does not change, and that patients are independent.

(a) How many patients on average should he expect to give the drug before one of them has a side effect?

(b) What is the probability that the 5th patient he gives the drug to is the first to have a side effect?

(c) What is the probability that the 2nd patient to have a side effect occurs after the 5th patient to receive the drug?

(d) What is the probability that the 2nd patient to have a side effect occurs exactly on the 13th patient to receive the drug?

Explanation / Answer

Given p =0.15

a) The number of patients on average should he expect to give the drug before one of them has a side effect = 1 / p = 6.66 = 7 approximatly

b) The probability that the 5th patient he gives the drug to is the first to have a side effect is

      P(X=5) = qk-1 p = (0.85)5-1(0.15) = 0.0783

c) The probability that the 2nd patient to have a side effect occurs after the 5th patient to receive the drug is

As Lack of Memory property(Markov Property), P(X=7/X>5) = P(X=2) = q2-1 p = (0.85)2-1(0.15) = 0.1275

d) The probability that the 2nd patient to have a side effect occurs exactly on the 13th patient to receive the drug

As Lack of Memory property(Markov Property), P(X=13/X>11) = P(X=2) = q2-1 p = (0.85)2-1(0.15) = 0.1275

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