Question

You are getting your wisdom teeth removed and are a participant in a clinicaltrial for three different procedures, one of which is randomly assigned to you. Of those whoundergo the first procedure, 8% get an infection; of those who undergo the second procedure,4% get an infection; of those who undergo the third procedure, 9% get an infection.

(a) What is the probability that you will not get an infection.

b) Unfortunately, you got an infection after removing your wisdom teeth! What is the probability that you were assigned to the first procedure?

Answer 1

(a)

I : Event of you will getting an infection

P1 : Event of you undergo First Procedure

P2 : Event of you undergo Second procedure

P3 : Event of you undergo Third Procedure.

You are getting your wisdom teeth removed and are a participant in a clinical trial for three different procedures, one of which is randomly assigned to you therefore

P(P1)=P(P2)=P(P3) = 1/3

Of those who undergo the first procedure, 8% get an infection i.e

Probability of getting infection given you undergo the first procedure : P(I|P1) = 8/100 =0.08

of those who undergo the second procedure,4% get an infection i.e

Probability of getting infection given you undergo the Second procedure P(I|P2) = 4/100 = 0.04

of those who undergo the third procedure, 9% get an infection i.e

Probability of getting infection given you undergo the Third procedure P(I|P3) = 9/100 = 0.09

Probability of getting an infection = P(I)

P(I) = 0.07

Probability that you will not get an infection :P() = 1 - P(I) = 1-0.07=0.93

Probability that you will not get an infection = 0.93

b) Unfortunately, you got an infection after removing your wisdom teeth! Probability that you were assigned to the first procedure = Probability that you were assigned to the first procedure given that you got an infection = P(P1|I)

By Bayes theorem,

From (a)

Unfortunately, you got an infection after removing your wisdom teeth! Probability that you were assigned to the first procedure = 0.380952381