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How many bulbs should we buy if we want to be 95% certain that we will have 1000 hours of light if lifetime time of each bulb is assumed to be exponentially distributed with mean $100(\frac{1}{\lambda})$ hours on the basis of the following conditions :

(1) Assuming all the bulbs are burning simultaneously :

Since all the bulbs are burning simultaneously , the way I see it , we should have :

$\sum_{i=1}^{n}( X_i) \geq 1000$ , thus ,

$P(\sum_{i=1}^{n}( X_i) \geq 1000)=0.95$

Now , $\sum_{i=1}^{n}( X_i)$~$Gamma(n,\lambda)$ and $(2\lambda \sum_{i=1}^{n}( X_i))$~ $Chi-Square(2n)$ , thus ,

$P(\sum_{i=1}^{n}( X_i) \geq 1000)=P(\chi_{2n}^{2} \geq 20)=0.95$ which gives $n \approx 16$.

(2) Assuming that one bulb is used until it burns out and then is replaced , etc.

Now since 1000 hours to be achieved , each bulb must have lifetime $\geq$ $\frac{1000}{n}$ , thus ,

$P(X_i \geq \frac{1000}{n})=0.95$ => $(-e^{- \lambda x_i})|_{\frac{1000}{n}}^{\infty}$ => $e^{- \lambda \frac{1000}{n}}$ => $e^{- \frac{10}{n}}=0.95$

which gives $n \approx 195$.

I don't know but that doesn't seem right , that much huge difference in the values on $n$ , can anyone help ?

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    I'm not sure I understand the difference between 1 and 2... For question 1, are you trying to find $ n $ such that there will be light after 1000 hours? I.e. you turn on $ n $ lights at time 0, and you want to know how many you need in order to be 95% confident that at least one of the lightbulbs is still working after 1000 hours? – 2017-01-16
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    The gist is , we want 1000 hours of light from $n$ number of bulbs , where $n$ we have to find. @MichaelR – 2017-01-16
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    How are questions 1 and 2 different? – 2017-01-16
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    Frankly speaking , seems same to me , but the text I am reading , it is given as two different cases ! @MichaelR – 2017-01-17

2 Answers 2

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For (1), $\sum_{i=1}^n X_i \geq 1000$ is too weak of a condition, because it allows, e.g., for the possibility that each bulb burns for $1000/n$ hours, which wouldn't yield 1000 hours of light. Instead, try computing the probability of getting fewer than 1000 hours of light. (What would this imply about each of the $X_i$ ?)

For (2), that isn't the condition you're looking for either. If $n = 2$, you're excluding, for instance, the possibility that Bulb 1 burns for 100 hours, and Bulb 2 burns for 900 hours. It'd actually be more appropriate to compute $P(\sum_{i=1}^n \geq 1000)$ here.

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    Okay . So for the (1) case I am thinking of using the joint distribution of $X_1,X_2$. . . . . . . ,$X_n$ , is that right ? – 2017-01-16
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    Yes, you could use the joint distribution, but how do you plan to use it? – 2017-01-16
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    A user has tried below ! – 2017-01-17
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Let $ X_1,..., X_n \sim Exponential\big(u=\frac{100}{\lambda}\big) $, which means that $ f_X{(x)} = \frac{1}{u} e^{-x/u} = \frac{\lambda}{100} e^{-x\lambda/100} $

Question 1: we want to find the probability that at least one lightbulb (out of $ n $) is still working after 1000 hours, i.e.

\begin{equation} P(\geq\text{1 lightbulb after 1000 hours}) = 1 - P(\text{0 lightbulbs after 1000 hours}) = 1 - \prod_{i=1}^{n}P(X_i \leq 1000) \\ = 1 - P(X \leq 1000)^n = 1 - \big(\int_{0}^{1000}\frac{\lambda}{100} e^{-x\lambda/100}\big)^n = 1 - \big(1 - e^{-1000\lambda/100}\big)^n = 0.95 \end{equation}

Solving for $ n $ as a function of $ \lambda $, we get:

\begin{equation} 1 - \big(1 - e^{-1000\lambda/100}\big)^n = 0.95 \rightarrow 0.05 = \big(1 - e^{-1000\lambda/100}\big)^n \\ \rightarrow log(0.05) = nlog\big(1 - e^{-1000\lambda/100}\big) \rightarrow n = \frac{log(0.05)}{1 - e^{-1000\lambda/100}} \end{equation}

Question 2: we want to find the probability that the total lifetime of $ n $ lightbulbs exceeds 1000 hours with 95% probability, i.e.

\begin{equation} P\big(\sum_{i=1}^{n}X_i \geq 1000\Big) = 0.95 \end{equation}