Let V be the event that a computer contains a virus, and let W be the event that a computer contains a worm. Suppose P(V) = 0.17 , P(W) = 0.05 , and P(V and W) = 0.04 . What is the probability that the computer contains neither a virus nor a worm?
Accepted Solution
A:
Answer: 0.82Step-by-step explanation:The probability of the computer not containing neither a virus nor a worm is expressed as P([tex]V^{C}[/tex]∩[tex]W^{C}[/tex]) , where P([tex]V^{C}[/tex]) is the probability that the event V doesn't happen and P([tex]W^{C}[/tex]) is the probability that the event W doesn't happen.P([tex]V^{C}[/tex])= 1-P(V) = 1-0.17 = 0.83P([tex]W^{C}[/tex])=1-P(W) = 1-0.05 = 0.95Since [tex]V^{C}[/tex] and [tex]W^{C}[/tex] aren't mutually exclusive events, then:P([tex]V^{C}[/tex]∪[tex]W^{C}[/tex]) = P([tex]V^{C}[/tex]) + P([tex]W^{C}[/tex]) - P([tex]V^{C}[/tex]∩[tex]W^{C}[/tex])Isolating the probability that interests us:P([tex]V^{C}[/tex]∩[tex]W^{C}[/tex])= P([tex]V^{C}[/tex]) + P([tex]W^{C}[/tex]) - P([tex]V^{C}[/tex]∪[tex]W^{C}[/tex])Where P([tex]V^{C}[/tex]∪[tex]W^{C}[/tex]) = 1 - 0.04 = 0.96Finally:P([tex]V^{C}[/tex]∩[tex]W^{C}[/tex]) = 0.83 + 0.95 - 0.96 = 0.82