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1 month ago

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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.

1 answer:

zzz [12.3K]1 month ago

8 0

Response:

$P(X = 0) = 0.42$

$P(X = 1) = 0.46$

$P(X = 2) = 0.12$

Detailed explanation:

We are presented with the following probabilities:

There is a 40% chance that the first file is corrupted. Thus, there is a 60% chance that the first file is intact.

The second file has a 30% chance of being corrupted, making the chance that it remains intact 70%.

Probability mass function

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

No files corrupted:

There is a 60% chance that the first file is fine.

Additionally, a 70% chance that the second file is also intact.

So

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

Exactly one file corrupted:

In this case, the first one is damaged, while the second remains functional.

This comprises a 40% chance for the first file being corrupted and a 70% chance for the second file being intact.

Alternatively,[the first remains fine with a 60% probability while the second gets corrupted (30%).

Two files corrupted:

A 40% probability applies to the first file being corrupted.

While the second file has a 30% chance of being compromised.

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

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3 months ago

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Provided data

Volume required $V=220\ in.^3$

For the initial cylinder

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For the second cylinder

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$V_4=452.448\ in^3$

Thus, cylinders $1$ and $2$ meet the requirement as they remain under $220\ in.^3$

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