Question

After a computer virus entered the system, a computer manager checks the condition of all important files. She knows that each file has probability 0.2 to be damaged by the virus, independently of other files. (a) Compute the probability that at least 5 of the first 20 files are damaged. (b) Compute the probability that the manager has to check at least 6 files in order to find 3 undamaged files

Answer 1

Answer with Step-by-step explanation:

The given probability can be obtained Bernoulli's probability theory since the outcome of the experiment is binary

For an event E with probability of occurrence 'p' the probability that the event  E occurs exactly 'r' times in 'n' trails is given by

$P(E)=\binom{n}{r}p^{r}(1-p)^{n-r}$

Applying the given values we get

since p = probability that the file has a virus = 0.2

n = no of trails = 20

r =  5 ( Since we need to find the probability that at least 5 files have virus)

Part a)

$P(E)=\binom{20}{5}(0.2)^{5}(1-0.2)^{20-5}\\\\P(E)=\frac{20!}{(20-5)!\cdot 5!}\times (0.2)^{5}(1-0.2)^{15}\\\\P(E)=0.1745$

Part b)

Let P(E') be the probability that the manager has to check 6 files to find 3 undamaged ones thus n = 6, r = 3

$P(E)=\binom{6}{3}(0.2)^{3}(1-0.2)^{6-3}\\\\P(E)=\frac{6!}{(6-3)!\cdot 5!}\times (0.2)^{3}(1-0.2)^{3}\\\\P(E')=0.082$