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You have a flashlight that uses two batteries and you have a package of three new batteries. Suppose that each battery has an independent lifetime (in hours) with exponential(1/50) distribution. Find the distribution of the number of hours T that you can use the flashlight.

Attempt for solution: denote by T the number of hours you can use the flash light let B1, B2 and B3 be the batteries respectively (B1\B2\B3) ~ exp(1/50)

I have the thought that if we use the first two batteries and the first one runs out first, then we have the second and third one once the second one runs out we can't use the flash light so the maximum life of the flashlight would be the amount of time the second battery runs

But i don't exactly know how to put this in terms of probability

I would really appreciate if there is someone out there willing to share ideas

Thanks alot!!

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I don't have an answer, but I think that strategy isn't optimal. Consider batteries with equal life $L$. Use B1 and B2 for $L/2$, swap B1 for B3, use those for $L/2$, swap B1 for B2, use those for $L/2$. Total: $3L/2$. I don't know the ideal time to swap the first time when it's a probability distribution. –  Henry Swanson Jun 29 '13 at 1:27
    
Are you comfortable with the distribution of a sum of two exponential random variables? –  André Nicolas Jun 29 '13 at 2:08
    
You should never mix old batteries and new batteries! It's a trick question! –  Brian Rushton Jun 29 '13 at 2:30

2 Answers 2

So the lifetime of each battery is an exponential distribution $X_{1,2,3}$ with $\lambda=0.02$ when in use and (unlike real batteries) they do not age when not in use.

So put 2 batteries in and the first will fail with $\min\{X_1,X_2\}$ which is exponentially distributed with $\lambda=0.04$.

When the first battery fails, put in battery 3. Because of the memoryless nature of exponential distributions, this is also exponential with $\lambda=0.04$ - but starting at a latter time!

So that gives us the sum of 2 i.i.d. exponential random variables with $\lambda=0.04$. This is an Earlang distribution with $k=2$ and $\lambda=0.04$. So the pdf is:

$$\begin{align}P(T=t) &=\frac{0.04^2te^{-0.04t}}{\Gamma(2)}\\ &=0.016te^{-0.04t}\\ \end{align}$$

See plot.

And the cdf is:

$$\begin{align}P(T\le t) &=\frac{\gamma(2,0.04t)}{(2-1)!}\\ &=\gamma(2,0.04t)\\ \end{align}$$

See plot.

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An alternative approach is to consider the lifetime of the flashlight to be the lifetime of any 2 of the batteries. The rationale is that the flashlight can only operate with 2 working batteries. Let X and Y be the random variables corresponding to the life of 2 of the batteries. Let T = X + Y. Then T represents the combined life of 2 of the batteries. Since T is the convolution of 2 exponential variables the p.d.f. is $f_T(t) = \lambda^2 t e^{-\lambda t}$. This corresponds to the $Gamma(n, \lambda)$ distribution which is defined as $$\frac{\lambda^n t^{n-1} e^{-\lambda t}}{(n-1)!}$$

where since $n-1$ is the number of spare batteries, $n=2$ in this case. So, since $f_T(t)$ is the gamma distribution, my understanding is that $F_T(t) = \int_0^t f_T(t) dt$ should be the distribution function. However, this not give the same answer as the original answer to this post. So, I'm obviously missing something. Could someone clarify where my approach went wrong?

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This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. –  amWhy Oct 29 '13 at 14:46
    
@andrenicolas, since you suggested the sum of 2 exponential variables, I've tried that approach and posted my solution to this post. However, I don't get the same solution as the original answer to this post. Could you tell me where I went wrong? Thx! –  EggHead Nov 1 '13 at 2:37

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