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 ) = 0.15, p(w) = 0.05, p(v ∪ w) = 0.17. (a) find the probability that the computer contains both a virus and a worm. (b) find the probability that the computer contains neither a virus nor a worm.

Answer 1

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 $P(v\cup w)$ is 0.17.

(a) To find out the probability that the computer contains both a virus and a worm we will use probability formula,

$P(A)+P(B)=P(A\cup B)+P(A\cap B)$

Now let us substitute our given values in this formula.

$0.15+0.05=0.17+P(A\cap B)$

$P(A\cap B)=0.20-0.17=0.03$

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.

$\text{Probability that computer contains neither virus nor worm}=1-P(v\cup w)$

$=1-0.17=0.83$

Therefore, probability that the computer contains neither a virus nor a worm is 0.83.