Answer: Our required probability is 87.12%.
Step-by-step explanation:
Since we have given that
P(infection occurs in operations ) = 2%
P(repair fails in operations ) = 12%
P(both infection and failure occur together) = 1.12%
We will first find P( infection or failure ) is given by
[tex]0.02+0.12-0.0112\\\\=0.1288\\\\=12.88\%[/tex]
So, Probability of these operations succeed and are free from infections is given by
[tex]100-12.88\\\\=87.12\%[/tex]
Hence, our required probability is 87.12%.