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Math Help - Probability 2 - Lightbulbs

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    Probability 2 - Lightbulbs

    Assume that a new light bulb will burn out after t hours, where t is chosen from [0,infinity) with an exponential density of f(t)= L*e^(-Lt)
    In this context, L is often called the failure rate of the bulb.

    a) assume that L = 0.01, and find the probability that the bulb will not burn out before T hours. This probability is often called the reliability of the bulb.
    b) for what T is the reliability of the bulb = 1/2?

    I set up part a as f(t) = 0.01e^(-0.01t), which is obvious, but I'm not sure where to go from here. Since this is the equation for the failure rate...would the probability of it NOT burning out be 0.99e^(-0.99t)? I am somewhat confused, my professor did not explain this too well.
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    Quote Originally Posted by mistykz View Post
    Assume that a new light bulb will burn out after t hours, where t is chosen from [0,infinity) with an exponential density of f(t)= L*e^(-Lt)
    In this context, L is often called the failure rate of the bulb.

    a) assume that L = 0.01, and find the probability that the bulb will not burn out before T hours. This probability is often called the reliability of the bulb.
    b) for what T is the reliability of the bulb = 1/2?

    I set up part a as f(t) = 0.01e^(-0.01t), which is obvious, but I'm not sure where to go from here. Since this is the equation for the failure rate...would the probability of it NOT burning out be 0.99e^(-0.99t)? I am somewhat confused, my professor did not explain this too well.
    (b) Solve for T:

    \Pr(\text{Lifetime of bulb} > T) = \frac{1}{2}, that is, \int_T^{+\infty} 0.01 e^{-0.01 t} \, dt = \frac{1}{2}.
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