Thread: Conditional Probability and Independence
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Old June 24th, 2008, 08:34 PM
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Quote:
Originally Posted by algebrapro18 View Post
Of all failures of a certain type of hard drive, it is determined that 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% of the cases both the allocation sector and one or more nonessential sectors are damaged. A failed device 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?

The teacher said to symbolize what we have and what is being asked of us. This will help solve the problem.

P(AS) = .2
P(NS) = .7
P(AS and NS) = .1

A. P(AS U (AS and NS)) = P(AS) + P(AS and NS) - P(AS and (AS and NS))
= .2 + .1 - ?
I don't know how to find that intersection

B. P(NS U (AS and NS)) = P(NS) + P(AS and NS) - P(NS and (AS and NS))
= .7 + .1 - ?
Again I don't know how to find that intersection

C. P(NS|AS) = P(NS and AS) / P(AS)
= .1 / .2 = .5

D. P(AS|NS) = P(AS and NS) / P(NS)
= .1 / .7 = .142857

So I need to know how to figure out those intersections in A and B and make sure C and D are right.
I'd suggest drawing a Karnaugh table:

\begin{tabular}{l | c | c | c} & (AS) & (AS)$\, '$ & \\ \hline NS & 0.1 & 0.7 & \\ \hline (NS)$\, '$ & 0.2 & & \\ \hline & & & 1 \\ \end{tabular}


Filling in the blanks should be easy:


\begin{tabular}{l | c | c | c} & (AS) & (AS)$\, '$ & \\ \hline NS & 0.1 & 0.7 & {\color{red}0.8} \\ \hline (NS)$\, '$ & 0.2 & {\color{red}0} & {\color{red}0.2} \\ \hline & {\color{red}0.3} & {\color{red}0.7} & 1 \\ \end{tabular}


Then, from the above table:

A. Pr(AS) = 0.3.

B. Pr(NS) = 0.8.

C. Pr(NS | AS) = 0.1/0.3 = 1/3.

D. Pr(AS | NS) = 0.1/0.8 = 1/8.
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