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The probability of a light bulb lasting T hours is $\exp{10}$ distributed:

$$f_T(t) = \frac{1}{10}e^{-t/10}$$

And the probability that the light bulb will be used H hours is Po(12) distributed:

$$p_H(k) = e^{-12}\cdot\frac{12^k}{k!}$$

I want to find the probability that the light bulb will last as week ($k=168$)

Therefore $P[T>H]$.

So from total probability I can get

$$P[T>H] = \sum_k P[T>H|H=k]P[H=k]$$

and therefore I want to find $\sum_k P[T>k]P[H=k]$ .... right?

If this is set up correctly, my main question is setting up the summations. I am quite rusty on them. I assume that I would first sum over $k=0$ to $k=168$, however, T would be from $0$ to $\infty$. I'm confused on how to proceed. Also I might need help with the computations.

Thanks

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What is $k$ in your Poisson distribution? –  Ron Gordon Feb 13 '13 at 18:13
    
$k$ = 0,1,2,3,... –  user61147 Feb 13 '13 at 18:33

1 Answer 1

Let $a=\frac1{10}$ and $b=12$. For every nonnegative $t$, $\mathbb P(T\gt t)=\mathrm e^{-at}$ hence, assuming that $T$ and $H$ are independent, $\mathbb P(T\gt H\mid H)=\mathrm e^{-aH}$ and $\mathbb P(T\gt H)=\mathbb E(\mathrm e^{-aH})$. For every $s$, $\mathbb E(s^H)=\mathrm e^{-b(1-s)}$, hence $\mathbb P(T\gt H)=\mathrm e^{-b(1-\mathrm e^{-a})}$. Numerically, $\mathbb P(T\gt H)$ is approximately $\mathrm e^{-ab}=\mathrm e^{-6/5}=0.301$ (exact value $0.319$).

Caveat: This takes things at the point where you declare that you are interested in the value of $\mathbb P(T\gt H)$. Unfortunately, I cannot discern how this quantity is related to "the probability that the light bulb will last as week" (whatever that means).

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