Answer:
The probability that the operation is successful and infection-free is 0.9831
Explanation:
Let the infections be denoted by I and the fails be denoted by F then
P(I)= 2% = 0.02
P (F)= 11% = 0.11
P(I∩F)= 0.5 % = 0.005
First we find the probability of infection or fail by using the general rule of addition
P (infection or repair) = P(I) +P (F) - P(I∩F)
= 0.02 + 0.11 -0.005
= 0.125
Now using the complement method to find the no infection or repair successful
P(no infection or repair successful)= 1- P (infection or repair)
= 1- 0.125= 0.875
P( Successful) = 1- P (F)= 1- 0.11= 0.89
The probability that the operation is successful and infection-free is given by the conditional probability
P( Successful given that no infection) = P(no infection or repair successful)/P(successful)
P( Successful given that no infection) = 0.875/0.89= 0.9831