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Homework Help: Probability semiconductor problem

  1. Feb 10, 2007 #1
    1. The problem statement, all variables and given/known data

    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 (A )
    -B (intersction) C

    2. Relevant equations

    Relative probability= n/N
    n- number of specific events occured.
    N- total number of posiible events

    3. The attempt at a solution

    for the number of events:
    I did 2^3=8 possible event but really I have no confindence in my answer
    For question 2 , I dont get it. I have formula but does nos help me much.

    Can I have some help from an expert please?
  2. jcsd
  3. Feb 10, 2007 #2
    I'm not an expert, but I thought your problem was interesting and had some free time, so I tried it. here's what I got.
    Each location is either contaminated or not and there are 6, so 2^6=64
    No contamination particles=0.6*.8*.85*.9*.95*.9
    This is just the product of the complements of the probabilities.
    Complement (A ) is just 1-0.6*.8*.85*.9*.95*.9

    Intersection B and C=This is the sum of probabilities 1 contamination particle+3 contamination particles

    First, let pro=0.6*.8*.85*.9*.95*.9 and sum=4/6+2/8+15/85+1/9+5/95+1/9 and sumsq=(4/6)^2+(2/8)^2... +(1/9)^2 and sumcu=(4/6)^3+(2/8)^3...+(1/9)^3

    Probability of 1 contamination particle is:

    Probability of 3 is a bit trickier:

    The /6 is to account for permutations. The -3*sum*sum^2 is to get rid of things such as 4/6*4/6*4/6 and 4/6*4/6*1/9, since one part cannot be contaminated twice. The +2sum^3 is because -3*sum*sum^2 over-subtracts the perfect cubes.

    Thus, B intersection C =pro*sum+pro/6*[sum^3-3*sum*sum^2+2*sum^3].

    I haven't done combinatorics for a while, so check my answer before you take my word for it.....

  4. Feb 10, 2007 #3
    Thank you sphoenixee,
    It is a little bit tricky but I get the idea.
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