What is the Probability of a Computer System Not Working with 3 Hard Disks?

[Maybe_Memorie]

Homework Statement

A computer system is equipped with 3 hard disks, each independantly having a 0.97 probability of functioning correctly. At least 2 of these are required for a working system. What is the probabiliy that the system is not working?

Homework Equations

P(A and B) = P(A) x P(B)
P(A or B) = P(A) + P(B)

The Attempt at a Solution

3 hard disks, A,B,C, each a 0.97 probability of working, 0.03 of not working.

P(not working) = 1 - p(working)
= p(A and B) or (A and C) or P (B and C)


What's wrong here? I'm getting a probability > 1.

 

[Filip Larsen]

The equations you gave are only valid for independent events.

For this problem, you may want to model the probability that exactly i hard disks are working, P(X=i), i = 0, 1, 2, 3, and use that to answer the original question.

 

[Maybe_Memorie]

Is this correct?

The probability it isn't working is the probability only 0 or 1 disks work.

P(X=0) = (0.03)(0.03)(0.03) = 0.000027

If we label the disks A,B,C, the prob that only 1 works = prob A works, B,C fail, and so on

P(X=1) = (0.97)(0.03)(0.03) + (0.97)(0.03)(0.03) + (0.97)(0.03)(0.03)
= 3(0.97)(0.03)(0.03)

Then the answer is P(X=0) + P(X=1)

Is this correct?

 

[Filip Larsen]

Yes, that is correct.

You may know the general solution as the Binomial distribution. You can find a brief description of it in most textbooks on probabilities or at http://en.wikipedia.org/wiki/Binomial_distribution.

 

[Maybe_Memorie]

I don't understand how this is Binomial

 

[Filip Larsen]

You can model the functioning of each hard disc as a "toss of a coin" that has 0.97 chance to land on "working" and 0.03 chance to land on "not working", i.e. the probability of a hard disc working can (in the context of this problem at least) be model as single event with a 0.97 probability of working. With 3 hard discs you then have 3 such independent events, each with the same probability of working.

If you look at the Binomial distribution, it gives the probability P(X = k), where X is a stochastic variable representing the number of successes when you have n independent events each having probability p for success. In your case, n is 3 and if we choose to model "success" as "hard disc is working", then p is 0.97, and the answer to the problem is P(K >= 2) = P(X = 2) + P(X = 3).

Notice, that since you can replace k with n-k and p with 1-p in the Binomial distribution and get same result, you may also map "success" in the Binomial distribution to "hard disc not working" with probability p = 0.03, and then you get the answer as P(K' < 2) = P(K' = 0) + P(K' = 1), where K' now represent the number of hard discs that doesn't work, i.e. K' = 3 - K. This last formulation corresponds to the solution you found in post #3.

 

[HallsofIvy]

The computer will NOT work if all three discs are not working. It is easy to find that probability. Then the probability is will work is 1 minus that value.

This is a "binomial" distribution because each disc has only two possible states, working and not working, just as a flipped coin has two possible states, heads or tails. You could have done this by saying the probability that the all three discs are working is (0.97)^3. And the probability that the first disc is not working but the other two are is (0.03)(0.97)^2 and that there are
\begin{pmatrix}3 \\ 1\end{pmatrix}= 3[/itex]<br /> ways of ordering the discs so the probability that exactly two are working is 3(0.03)(0.97)^3. And, finally, the probability that the first two are not working but the third is is<br /> \begin{pmatrix}3 \\ 2\end{pmatrix}(0.03^2)(0.97)= 3(0.03^2)(0.97)<br /> and so the probability that all three, or two, or one, of the discs is working is<br /> (0.97)^3+ 3(0.97^2)(0.03)+ 3(0.97)(0.03^2)<br /> <br /> However, the method in my first paragraph is simplest and will give the same answer.

 

[Filip Larsen]

The computer will NOT work if all three discs are not working. It is easy to find that probability.

Three non-working discs is not the only case for which the computer won't work. If two of the three discs aren't working the computer won't work either.

However, the method in my first paragraph is simplest and will give the same answer.

No, it won't. One may think that if you had actually tried it, you would have caught your error before posting (sorry, couldn't resists teasing you with this )