Question

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?

Answer 1

Answer: 0.82

Explanation:

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

P(
`V^(C)`)= 1-P(V) = 1-0.17 = 0.83

P(
`W^(C)`)=1-P(W) = 1-0.05 = 0.95

Since
`V^(C)` and
`W^(C)` aren't mutually exclusive events, then:

P(
`V^(C)`∪
`W^(C)`) = P(
`V^(C)`) + P(
`W^(C)`) - P(
`V^(C)`∩
`W^(C)`)

Isolating the probability that interests us:

P(
`V^(C)`∩
`W^(C)`)= P(
`V^(C)`) + P(
`W^(C)`) - P(
`V^(C)`∪
`W^(C)`)

Where P(
`V^(C)`∪
`W^(C)`) = 1 - 0.04 = 0.96

Finally:

P(
`V^(C)`∩
`W^(C)`) = 0.83 + 0.95 - 0.96 = 0.82