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Math Help - probability 2

  1. #1
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    probability 2

    Hi, I have some issues with another problem.
    Here it is:

    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)
    -B union C


    For the first question:
    Total events =2^6

    For the second question I did:

    probability for No contamination particles=0.6*.8*.85*.9*.95*.9

    so , Complement (A ) is just 1-(0.6*.8*.85*.9*.95*.9)
    I dont know if I get it right
    Now for the rest I get lost...

    Please tell me If I got the first two questions.

    thank you
    B
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  2. #2
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    Lexington, MA (USA)
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    Hello, braddy!

    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 0.40, 0.20, 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 (B intersection C)
    . . .B union C


    For the first question:
    Total events = 2^6 . No, there are simply six possible outcomes.

    For the second question I did:

    Probability for No contamination particles = (0.6)(0.8)(0.85)(0.9)(0.95)(0.9)

    so: Prob(A') is just: 1 - (0.6)(0.8)(0.85)(0.9)(0.95)(0.9) . Right!

    Prob(B ∩ C) means: at most 3 particles and an odd number of particles.
    . . That is, there are 1 or 3 particles.
    The probability is: .0.2 + 0.1 .= .0.3


    Prob(B U C) means: at most 3 particles or an odd number of particles.
    . . That is, there are 0, 1, 2, 3 or 5 particles.
    The probability is: .0.4 + 0.2 + 0.15 + 0.10 + 0.10 .= .0.95

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