Question

A computer virus is trying to corrupt two files. The first file will be corrupted with probability 0.4. Independently of it, the second file will be corrupted with probability 0.3. Compute the PMF of X, the number of corrupted files.

A = probability of corrupted first file
B = probability of corrupted second file

So,
for X = 2, P(A and B) = 0.4 * 0.3 = 0.12
for X = 0, P(~A and ~B) = 0.6 * 0.7 = 0.42
for X = 1, P(A or B) = 0.4 + 0.3 - 0.12 = 0.58 ??? but it should be 0.46, what is wrong with my reasoning?

Answer 1

for one file:


For X = 1, P(~A and B)or P(A and ~B) = 0.6 * 0.3 + 0.4*0.7 = 0.46

Answer 2

Please confirm if the above reasoning is correct.
i.e. for only one file implies that one file is affteced and not the other.
i.e either
A is affected and B is not affected
or
B is affected and A is not affected

Answer 3

Oh, I think I see, the universe for P(A and B) is first or second file while P(~A and B) or P(A and ~B) the universe is first, second or no file.

Answer 4

yes I think you are right

Answer 5

Thanks!

Answer 6

vw