Question

ann works as an it technician for a local company. there are 6 computers infected with a virus on the company’s network of 1000 computers. ann chooses a computer on the company’s network at random. let the event a and the event b be as follows. a: the computer ann chooses is infected with the virus. b: the computer ann chooses is not infected with the virus. find the following probabilities. write your answers as decimal numbers and do not round. p(a)= p(b)=

Explanation:

Step1: Recall probability formula

The probability of an event $E$ is given by $P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.

Step2: Calculate $P(A)$

The number of favorable outcomes for event $A$ (infected computers) is $6$, and the total number of computers is $1000$. So $P(A)=\frac{6}{1000}= 0.006$.

Step3: Calculate $P(B)$

The number of non - infected computers is $1000 - 6=994$. So $P(B)=\frac{994}{1000}=0.994$.

Answer:

$P(A)=0.006$
$P(B)=0.994$