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. (a) Compute the probability mass function (pmf) of X, the number of corrupted files.

Answer 1

Answer:

$P(X = 0) = 0.42$

$P(X = 1) = 0.46$

$P(X = 2) = 0.12$

Step-by-step explanation:

We have these following probabilities:

40% probability that the first file is corrupted. So 60% probability that the first file is not corrupted.

30% probability that the second file is corrupted. So 70% probability that the second file is not corrupted.

Probability mass function

Probability of each outcome(0, 1 and 2 files corrupted).

No files corrupted:

60% probability that the first file is not corrupted.

70% probability that the second file is not corrupted.

So

$P(X = 0) = 0.6*0.7 = 0.42$

One file corrupted:

First one corrupted, second no.

40% probability that the first file is corrupted.

70% probability that the second file is not corrupted.

First one ok, second one corrupted.

60% probability that the first file is not corrupted.

30% probability that the second file is corrupted.

$P(X = 1) = 0.4*0.7 + 0.6*0.3 = 0.46$

Two files corrupted:

40% probability that the first file is corrupted.

30% probability that the second file is corrupted.

$P(X = 2) = 0.4*0.3 = 0.12$