1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations



    3. 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..
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?