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Basic probability

  1. Jan 16, 2017 #1
    1. The problem statement, all variables and given/known data
    Untitled.png

    2. Relevant equations


    3. The attempt at a solution
    I can't understand the solution.
    Shouldn't the probability be
    0.99^2 x (0.96^2 + 0.96(0.04) + 0.04^2) + (0.01 (0.99) +0.01^2) x (0.96^2).
    Please someone explain the solution to me
     
  2. jcsd
  3. Jan 16, 2017 #2

    BvU

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    Hi,

    So what are the relevant equations for (a) and (b) ?
    Which one? There are two solutions !
     
  4. Jan 16, 2017 #3
    I don't understand part a. I have no idea.
    I think the relevant equations for (a) and (b) are simply basic probability rules like P(A) X P(B) = P(A and B)
    thanks
     
  5. Jan 16, 2017 #4

    Stephen Tashi

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    You didn't explain your own solution. I can't tell what you are thinking.

    The text is computing:
    Pr( lot accepted) =
    Pr(lot is from B)Pr( two working fuses are selected| lot is from B) + P(lot is from A) Pr( two working fuses are selected from lot | lot is from A).

    The wording of the problem does not tell us that each bath of fuses from manufacturer A contains exactly 4 defective fuses. So we must treat the probability that a "randomly selected fuse" from manufacturer A is defective as .04. By contrast if we knew that each batch of fuses from A contained exactly 4 defective fuses, we would be faced with a situation of "random sampling without replacement".

    The wording of the problem says that a given "lot" of fuses consists of fuses from only one manufacturer, not fuses from both manufacturers. The numbers you used in your solution seem to come from considering fuses taken from both manufacturers during one test for acceptance.
     
  6. Jan 16, 2017 #5

    BvU

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    So what are P(A) and P(B) ? How do you think this testing takes place ? And how do you explain your 0.99^2 x (0.96^2 + 0.96(0.04) + 0.04^2) + (0.01 (0.99) +0.01^2) x (0.96^2) with this relevant equation ?
     
  7. Jan 16, 2017 #6
    I interpret the question as taking 2 fuses from each of the box. When the 2 fuses from one box are not defective, the company accept one box. So I thought question a is asking the probability of accepting one box only (not both box accepted nor both not accepted). Now I understand the question is asking the probability of the company accepting one lot when only one box is examined. Thank you so much for both answers!
     
  8. Jan 16, 2017 #7

    Ray Vickson

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    The wording of (a) asks about a box; that means: (i) we receive a box; (ii) we draw two items from that box; (iii) we accept the box if both tested items are OK. So:
    $$P(\text{accept}) = P(A)\, P(\text{accept}|A) + P(B)\, P(\text{accept}|B).$$
     
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