Let V be the event that a computer contains a virus, and let W be the event that a computer contains a worm. Suppose =PV0.48, =PW0.28, PV and =W0.17.(a) Find the probability that the computer contains either a virus or a worm or both.(b) Find the probability that the computer does not contain a virus.
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 =PV0.48, =PW0.28, PV and =W0.17.(a) Find the probability that the computer contains either a virus or a worm or both.(b) Find the probability that the computer does not contain a virus.
Solution
(a) The probability that a computer contains either a virus or a worm or both can be found using the formula for the union of two events: P(V ∪ W) = P(V) + P(W) - P(V ∩ W). Substituting the given values, we get P(V ∪ W) = 0.48 + 0.28 - 0.17 = 0.59.
(b) The probability that a computer does not contain a virus is the complement of the probability that it does contain a virus. This can be found using the formula P(not V) = 1 - P(V). Substituting the given value, we get P(not V) = 1 - 0.48 = 0.52.
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