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?

0.82

Step-by-step explanation:

The probability of the computer not containing neither a virus nor a worm is expressed as P(∩) , where P() is the probability that the event V doesn't happen and P() is the probability that the event W doesn't happen.

P()= 1-P(V) = 1-0.17 = 0.83

P()=1-P(W) = 1-0.05 = 0.95

Since and aren't mutually exclusive events, then:

P(∪) = P() + P() - P(∩)

Isolating the probability that interests us:

P(∩)= P() + P() - P(∪)

Where P(∪) = 1 - 0.04 = 0.96

Finally:

P(∩) = 0.83 + 0.95 - 0.96 = 0.82