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A person has $100$ light bulbs whose lifetimes are independent exponentials with mean $5$ hours. If the bulbs are used one at a time, with a failed bulb being replaced immediately by a new one, approximate the probability that there is a working bulb after $525$ hours.

When I tried to solve the question, I tried to find $P\{X\geq1\}$, where $X$ is the number of working bulbs after $525$ hours, and $X = \sum_{i=1}^{100} X_i$ where $X_i$ is an indicator variable for a bulb whose lifetime is greater than $525$ hours. If the bulb has a lifetime greater than $525$ then its value is $1$, otherwise $0$. I then used the central limit theorem with mean $= 5$ and variance $= 5$. My answer seems to be incorrect.

The solution manual says that $X$ must represent the total lifetime of all the bulbs, and we must find $P\{X\geq525\}$ using the central limit theorem. I don't understand how the total lifetime of all the bulbs (sum of all the lifetimes) being greater than $525$ means that there is a working lightbulb after $525$ hours. Can't the sum of the lifetimes of all the bulbs be greater than $525$ even if the lifetime of each bulb is less than $525$?

Please help me resolve my confusion!

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“...there is a working bulb after $525$ hours” means that the sum of lifetimes of all $100$ bulbs exceeds $525$ hours and not that an individual bulb had that lifetime. And the sum of the lifetimes may exceed $525$: after the first bulb failes, it is immediately replaced by a new one, if that second one failes by a third one and so on. Hence we want to calculate $P(X_1+\cdots X_{100})>525$.

As the expected value of $X_i$ is $5$ and variance is $25$ we know that the expected value of $X$ is $100\cdot5$ and its variance is $100\cdot 25$. Now $X$ is approximately normal distributed, just calculate $P(X>525)$.

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The exercise seems assumes, without really saying it in so many words, that a bulb is only at risk of failing once you turn it on. So the exponentially distributed lifetime of each bulb only begins running after the previous bulb has failed.

Therefore the time it takes for all the bulbs to fail is the sum of the failure times of each of the bulbs.

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