We have been given that the probability of a computer containing a virus P(v) is 0.15 and probability of computer containing a worm P(w) is 0.05.
We are also told that probability of computer containing either virus or worm [tex]P(v\cup w)[/tex] is 0.17.
(a) To find out the probability that the computer contains both a virus and a worm we will use probability formula,
[tex]P(A)+P(B)=P(A\cup B)+P(A\cap B)[/tex]
Now let us substitute our given values in this formula.
[tex]0.15+0.05=0.17+P(A\cap B)[/tex]
[tex]P(A\cap B)=0.20-0.17=0.03[/tex]
Therefore, probability that that the computer contains both a virus and a worm is 0.03.
(b) To find out the probability that the computer contains neither a virus nor a worm we will subtract the probability of computer containing both virus and worm from 1.
[tex]\text{Probability that computer contains neither virus nor worm}=1-P(v\cup w)\\[/tex]
[tex]=1-0.17=0.83[/tex]
Therefore, probability that the computer contains neither a virus nor a worm is 0.83.
