Question

The director of an insurance company’s computing center estimates that the company’s computer has a 20% chance of catching a computer virus. However, she feels that there is only a 6% chance of the computer’s catching a virus that will completely disable its operating system. If the company’s computer should catch a virus, what is the probability that the operating system will be completely disabled?

Answer 1

Let $A$ denotes the event "catch a computer virus", $B$ denotes the event "virus completely disables operating system".

Given $P(A)=0.20,$ $P(A\cap B)=0.06.$

$$P(B|A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{0.06}{0.2}=0.3$$

If the company’s computer should catch a virus, the probability that the operating system will be completely disabled is $0.3.$