Of all failures of a certain type of computer hard drive, it is determined that in 20% of them only the sector containing the file allocation table is damaged, in 70% of them only nonessential sectors are damaged, and in 10% the cases both the allocation sector and one or more nonessential sectors are damaged. A failed drive is selected at random and examined.

a. What is the probability that the allocation sector is damaged?

b. What is the probability that a nonessential sector is damaged?

c. If the drive is found to have a damaged allocation sector, what is the probability that some nonessential sectors are damaged as well?

d.If the drive is found to have a damaged nonessential sector, what is the probability that the allocation sector is damaged as well?

e.If the drive is found to have a damaged allocation sector, what is the probability that no nonessential sectors are damaged?

f. If the drive is found to have a damaged nonessential sector, what is the probability that the allocation sector is not damaged?

Step 1 of 8:

It is given that a computer hard drive consists of two sectors,sector containing the file allocation table and the nonessential sector.

Also it is given that in 20% of failed hard drives,only the sector containing the file allocation table is damaged,in 70% of the failed hard drives only the nonessential sectors are damaged and in 10% of the failed hard drives both the allocation sectors and nonessential sectors are damaged.

Let us define event A as sector containing the file allocation table is damaged and event B as the nonessential sector is damaged.

A randomly chosen drive is examined.

Using all these we have to find the required probabilities.

Step 2 of 8:

We have,

P(A=0.20

P(B)=0.70

P(AB)=0.10

where,represents the event that the sector containing the file allocation sector is not damaged, and represents the event that the nonessential sector is not damaged.

Step 3 of 8:

(a)

Here we have to find the probability that the allocation sector is damaged.

That is,we have to find P(A).

P(A) is given by,

P(A)=P(A)+P(AB)

=0.20+0.10

=0.30

Thus, the probability that the allocation sector is damaged is 0.30.

Step 4 of 8:

(b)

Here we have to find probability that the nonessential sector is damaged. That is we have to find P(B).

It is given by,

P(B)=P(+P(AB)

=0.70+0.10

=0.80

Thus, the probability that the nonessential sector is damaged is 0.80.