probability

Discussion in 'Math' started by braddy, Feb 10, 2007.

  1. braddy

    Thread Starter Well-Known Member

    Dec 29, 2004
    83
    0
    Hi, Please can I have some help with problem.

    The following is a random experiment.
    A wafer from a semiconductor manufacturing is to be selected randomly and a location on the wafer inspected for contamination particles. The sample space for the number of contamination particles at the inspected location is S= {0, 1, 2, 3, 4, 5}.

    Relative frequencies for these outcomes are 0.4, 0.2, 0.15, 0.10, 0.05 and 0.10 respectively.
    Use relative as probabilities.
    Let A be the event that there are no contamination particles at the inspected location.
    Let B be the event that there are at most three contamination particles at the inspected location.
    Let C be the event that there are an odd number of contamination particles at the inspected location.
    1- How many events are possible?
    2- Find the probability of the following:
    - complement of A
    - B and C ( or B intersection C)

    For the first question:
    Total events =2^6

    For the second question I have no idea.

    thank you
    B
     
  2. omnispace

    Member

    Jul 25, 2007
    27
    0
    First let's define A, B, and C using set notation:
    A = {0}
    B = {3, 4, 5}
    C = {1, 3, 5}

    The complement of A, A' = {1, 2, 3, 4, 5}, meaning there at least one particle.

    The probability of A', P(A') = 1 - P(A) = 0.6.

    B intersect C means including the events that occur in BOTH B and C, so...
    B and C = {3, 5}.

    P (B and C) = P(3) + P(5) = 0.2.


    Er, I just noticed the date on the original post, but hope this helps somebody anyway.
     
  3. kautilya

    Active Member

    Apr 26, 2007
    39
    0
    Let us define the individual spaces:
    A={0}
    B={0,1,2,3} (at most three contaminated particles}
    C={1,3,5}

    P(A)=0.4
    Complement of A= 1-0.4 = 0.6

    P(B intersection C) = P(B)P(C) = 0.3
     
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