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

Step-by-step 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