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?

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?

anonymous · Answer

please explain this question in detail step by step

apoorvk · Answer

'P' of either A or B being corrupted, can be seen in two ways: (A)(B~) and (A~)(B).. the sum of these two probabilities is the total probability of ANY ONE of the two files getting infected.(if one gets infected we also have to include the calculation for the part that the other is NOT infected)

what you did right there was that you found the probability of ATLEAST one file being corrupted, which means that both files could also be infected.

Ans(c) P[(A)(B~)] + P[(A~)(B)] = 0.4*0.7 - 0.6*0.3 = 0.46

anonymous · Answer

thankew