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Math Help - I am not fond of riddles.

  1. #1
    Member billym's Avatar
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    I am not fond of riddles.

    A room has three air-conditioners: A, B, and C. For each of the them the times that they will work before they breakdown is exponentially distributed with mean 10 hours. Initially A and B are turned on and C is left to replace whichever of A or B dies first. When there is only one left working, what is the probability that is is C?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by billym View Post
    A room has three air-conditioners: A, B, and C. For each of the them the times that they will work before they breakdown is exponentially distributed with mean 10 hours. Initially A and B are turned on and C is left to replace whichever of A or B dies first. When there is only one left working, what is the probability that is is C?
    It is 0.5.

    Since the process has no memory once one breaks down we are left with the other and C, and both of their remaining lives are exponentially distributed with mean 10 hours - so it is a simple lottery, there is no distinction between the air conditioners so they are equally likely to be the first to break down.

    CB
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    Member billym's Avatar
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    But surely if the survivor of A or B has been running for the amount of time that it takes for the other to breakdown, wouldn't it be near death itself? How could it be equally as likely to break down as C which has started fresh?

    Isn't that like saying a machine is just as likely to breakdown in 20 hours as it is in 10?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by billym View Post
    But surely if the survivor of A or B has been running for the amount of time that it takes for the other to breakdown, wouldn't it be near death itself? How could it be equally as likely to break down as C which has started fresh?

    Isn't that like saying a machine is just as likely to breakdown in 20 hours as it is in 10?
    That is exactly the property of the exponential distribution, if it has been running for time T it is just as likely to break down in the next second as it was in the first second, that's what I meant by memoryless process. You can prove this yourself, you just have to compute the conditional distribution of survival time given the item has already survived a time T, you will find this it is the same as the distribution of survival times you started with.

    It is the same as saying that the machine is just as likely to break down between hours 10 and 20 given that it has not broken down already as it was to have broken down between hours 0 and 10.

    CB
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