Answer:
84.78%
Step-by-step explanation:
Data provided in the question:
P( Infection occurred ) = 4% = 0.04
P( Repair fails ) = 12% = 0.12
P( Both infection and failure occur ) = 0.78% = 0.0078
Now,
P( operations succeed and are free from infection )
= 1 - P( Failure ∪ Infection )
also,
P( Failure ∪ Infection )
= P( Infection occurred ) + P( Repair fails ) - P( Both infection and failure occur )
or
P( Failure ∪ Infection ) = 0.04 + 0.12 - 0.0078 = 0.1522
Therefore,
P( operations succeed and are free from infection )
= 1 - P( Failure ∪ Infection )
or
= 1 - 0.1522
= 0.8478
or
⇒ 0.8478 × 100% = 84.78%