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Math Help - Light bulb

  1. #1
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    Light bulb

    A room has two lamps that use bulbs of type A and B, respectively.The lifetime, X, of any particular bulb of a particular type is a random variable, independent of everything else, with the following PDF:


    for type-A Bulbs: fX(x) =
    e^−x, if x ≥ 0,
    0, otherwise;


    for type-B Bulbs: fX(x) =
    3e^−3x, if x ≥ 0,
    0, otherwise.


    Both lamps are lit at time zero. Whenever a bulb is burned out it is immediately
    replaced by a new bulb.
    (a) What is the expected value of the number of type-B bulb failures until time t?


    The time is infinity? how is this possible?


    (b) What is the PDF of the time until the first failure of either bulb type?


    lamdatotal = lamda 1 + lamda2
    4te^(-4t) ?


    (c) Find the expected value and variance of the time until the third failure of a
    type-B bulb.


    ???
    (d) Suppose that a type-A bulb has just failed. How long do we expect to wait until a subsequent type-B bulb failure?
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  2. #2
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    Re: Light bulb

    (a) Expected number of lightbulb failures would be infinity as well if t is allowed to range to infinity (assuming you have an infinite number of bulbs). I suspect they mean a general "t", and this is just a Poisson Process where you are expected to find E[N(t)], where N(t) is the number of bulbs busted by time "t".

    (b) Not. . .quite. You are looking for P(A or B < t), which is simply 1-P(A and B > t). Using independence of the processes you should be able to go from here (by relating the CDF to the PDF).

    (c) Use (a) with t=3.

    (d) Use independence and memoryless property of an exponential rv.
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  3. #3
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    Re: Light bulb

    a) E[x]=3t
    b) 1-(e^-x + 3e^-3x) ?
    c) This is for time? not number of arrivals? is it just E[x]=3/3 and V[x]=1/3
    d) E[x]=1/3
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  4. #4
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    Re: Light bulb

    On (b), P(A and B > t) = P(A > t)P(B >t). There shouldn't be addition anywhere in there. On (c), you are correct - I read that wrong. You can define a new random variable Z = X1+X2+X3, which will give the sum of the failure times of the first three bulbs (i.e. time until 3rd bulb fails). Using properties of the sum of exponential (which is another distribution you should know) you should be able to answer this quite quickly (assuming you didn't do that to arrive at the answer above).
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  5. #5
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    Re: Light bulb

    b) 1-(e^-x * 3e^-3x) = 1-4e^-4x ?? whats with the minus 1? 1-P(a and b) = p(no fail?)
    c) I used the formula E[x] = k/lamda and V[x] = k/lamda, where k = #arrivals? is this okay to do that?
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