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Probability Syntax

  1. Feb 3, 2012 #1
    A certain system can experience three different types of defects. Let A (i=1,2,3) denote the event that the system has a defect of type i. Suppose that

    P(A[1]) =.12
    P(A[2])=.07
    P(A[3])=.05
    P(A[1] union A[2])=.13
    P(A[1] union A[3])=.14
    P(A[2] union A[3])=.10
    P(A[1] intersects A[2] intersects A[3])=.01

    1.what is the prob that the system does not have a type one defect?

    2.what is the prob that the system has both type 1 and type 2 defects?

    3. What is the prob that the system has both type 1 and type 2 defects but not a type 3 defect?

    4.What is the prob that the system has at most two of these defects?


    I know #1 is .88 and #4 is .99, but I am having difficulty understanding #2 and #3.

    For #2: What is the probability that the system has both type 1 and type 2 defects, could that also include a system with ALL the defects (type 1, type 2, AND type 3 defects)? If so, I calculate the probability to be .07:

    ==>P(A int B) + P(A int B int C)
    ==>.12 + .07 - .13 + .01
    ==>.07

    If so, then then the probability of #3 is .06.

    Let me know if this is right or wrong.
     
  2. jcsd
  3. Feb 4, 2012 #2

    Stephen Tashi

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    Science Advisor

    Yes. The event [itex] A_1 \cap A_2 [/itex] has the event [itex] A_1 \cap A_2 \cap A_3 [/itex] as a subset.

    By what you said above, you don't have to add the probability of (A int B in C) to the probability of A int B. The probability of A int B already accounts for the probability of A int B int C.

    Also, I don't know how you calculated P(A int B).

    Use the equation [itex] P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2) [/itex] and solve it for [itex] P(A_1 \cap A_2) [/itex].
     
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