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Math Help - The event composition problem

  1. #1
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    The event composition problem

    Suppose that two defective refrigerators have been included in a shipment of six refrigerators. The buyer begins to test the six refrigerators one at a time.

    a) What is the probability that the last defective refrigerator is found on the fourth test?

    b) What is the probability that no more that four refrigerators need to be tested to locate both of the defective refrigerators?

    c) When given that exactly one of the two defective refrigerators has been located in the first test, what is the probability that the remaining defective refrigerator is found in the third or fourth test?

    Don't know where to begin here =(...
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  2. #2
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    If the refrigerators are denoted by GGGGBB: four good ones and two bad.
    There are 15 ways to rearrange that string.

    a) BGGBGG, GBGBGG, GGBBGG represent the only strings in which the last defective refrigerator is found on the fourth test. So what is the probability?

    b) What strings represent “no more that four refrigerators need to be tested to locate both of the defective refrigerators”?

    See if you can do part c.
    Last edited by Plato; January 30th 2008 at 12:27 PM.
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  3. #3
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    Smile

    a) 1/5?

    b) BBGG GG; BGBG GG; BGGB GG; GGBB GG; so 4/15?

    c) BGGB GG; BGBG GG so 2/15?

    Thank you again =) I guess I have overestimated the difficulty of this one, I was going in a totally different complicated direction to solve it.
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  4. #4
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    For (b): BBGG, BGBG, BGGB, GBGB, GBBG, GGBB
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  5. #5
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    Hello again, coming back to this problem. There is another solution that is bugging me now.

    Let:
    P(defective found) = P(D) = 1/3
    P(nondefective found) = P(N) = 2/3

    a) Probability that the last defective refrigerator is found on
    the fourth test = P(NDND)+P(NNDD)+P(DNND)
    = 3*(1/3)^2*(2/3)^2
    =0.1481



    b) Probability that no more that four refrigerators need to
    be tested to locate both of the defective refrigerators
    = P(DD) +P(DND)+P(NDD)+0.1481
    =(1/3)^2+2*(2/3)*(1/3)^2+0.1481
    =0.4074


    c) The required Probability = P(DND)+P(DNND)
    = (2/3)*(1/3)^2+(1/3)^2*(2/3)^2
    =0.1235


    Since the question is in fact saying that buyer is testing 1 fridge at a time, could this be the right sol?
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  6. #6
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    I got it now, the last solution mentioned by me is only correct when we have independent events, in this fridge case events are dependent, so can't use it.
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