'Any event' and 'more than one event' question

[zeion]

Homework Statement

Q:A school has a computer lab that is closed for the summer, when no students are there, and then is re-opened in the fall. Sometimes, a computer in the lab that worked before the lab closed is found to not work properly when it is powered up again in the fall.
These computer failures are always due to one or more of the following causes:
A) Failure of the power supply
B) Failure of the disk drive
C) Failure of the RAM
D) Failure of the CPU
From long experience, it is known that these failures events are independent, and that their probabilities of occurrence (for a single computer) are P(A) = 0.04, P(B) = 0.03, P(C) = 0.02, P(D) = 0.01.

1) Find the probability that a computer will fail for any of these reasons - that is, find P(F), where F = A∪B∪C∪D.

2) Find the probability that a non-working computer has failed for more than one reason - that is, find the conditional probability that one of A, B, C, and D has occurred, given that F = A ∪B∪C∪D has occurred.

Homework Equations

The Attempt at a Solution

1) P(F) = P(A∪B∪C∪D) = P(A) + P(B) + P(C) + P(D) - P(A∩B) - P(A∩C) - P(B∩D) - P(C∩D) + P(A∩B∩C∩D)
= 0.09750025

2) Let this event be called E. Find P(E | F) = P(E∩F) / P(F). But E is a subset of F so P(E∩F) = P(E).
P(E) is anything other than just A or just B.. etc.. so
P(E) = P((A∩B)∪(A∩C)∪(B∩D)∪(C∩D)) .. and I have to expand that again with the formula in 1)?
Not sure..

 

[mighty2000]

I would like to add some additional information and suggestions to this forum post. Firstly, it is important to note that these probabilities are based on past experience and may not necessarily reflect the current situation. It is always a good idea to regularly check and maintain the computers during the summer break to prevent any potential failures.

Additionally, it would be helpful to have a system in place to identify and record which computers have failed and for what reasons. This can help in identifying any patterns or trends in the failures and can aid in troubleshooting and preventing future failures.

In terms of finding the probability of a non-working computer having failed for more than one reason, it would be beneficial to calculate the individual probabilities of each combination of failures (e.g. P(A∩B), P(A∩C), etc.) and then add them together. This can give a more accurate estimate compared to assuming that all combinations have the same probability.

Lastly, it may also be worth considering implementing a regular maintenance schedule for the computers to ensure they are in good working condition when the lab re-opens in the fall. This can help prevent any unexpected failures and ensure a smooth start to the school year.