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October 3rd, 2016, 01:23 PM  #1 
Newbie Joined: Oct 2016 From: New Jersey Posts: 2 Thanks: 0  Uniform Probabilty
I need help in solving this problem. Problem The lifetimes of batteries are independent uniformly distributed random variables over (0,1) month. A device requires 2 batteries to operate. when a battery fails, both batteries in service are replaced and discarded. If you have a total of n batteries, find the mean of the total amount of time that the device can operate. 
October 3rd, 2016, 03:16 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,406 Thanks: 1307 
first thing I'd do is treat pairs of batteries as a single entity. $Pr[\text{battery pack life}<t]=Pr[\text{battery 1 life}<t]Pr[\text{battery 2 life}<t] = t^2$ so the pdf of a battery pack lifetime is the derivative of this, i.e. $p(t) = 2t,~t \in [0,1]$ Since we always replace an entire pack we'll consider first $n$ being even so there are $\dfrac n 2$ packs. The lifetime of the device is just the sum of the lifetimes of these $\dfrac n 2$ packs. Each pack has an expected lifetime of $E[t] = \displaystyle{\int_0^1}2t^2~dt = \dfrac 2 3$ and the expected value of this sum is just the sum of the expected values so $E[t] = \dfrac n 2 \dfrac 2 3 = \dfrac n 3$ If $n$ is odd, then the battery left over can't be used for anything and we essentially have $n1$ batteries to make $\dfrac {n1}{2}$ packs. In this case $E[t] = \dfrac{n1}{3}$ So our final expectation is $E[t] = \dfrac{\left \lfloor \frac n 2 \right \rfloor}{3}$ 
October 3rd, 2016, 06:37 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,406 Thanks: 1307 
The solution in post #2 is incorrect. The following is correct. Again treat two batteries at a time as a single entity called a pack. $Pr[\text{life of pack}<t] =1Pr[\text{life of pack}>t]^2$ In other words in order for the pack to have failed by time $t$ both batteries cannot have lasted longer than $t$ $Pr[\text{life of pack}>1] = 1t$ $Pr[\text{life of pack}<t]=1(1t)^2 = 2tt^2$ Thus the pdf of the pack lifetime is $p(t) = \dfrac{d}{dt} 2tt^2 = 2(1t)$ The expected lifetime of a pack is thus $E[t] = \displaystyle{\int_0^1}2t(1t) = \dfrac 1 3$ and again the Expected value of the sum of $\dfrac n 2$ of these packs is just $E\left[t  \dfrac n 2 \text{ packs }\right] = \dfrac n 2 \dfrac 1 3 = \dfrac n 6$ and by the argument of post #2 the final answer is $E[t  n \text{ batteries }]=\dfrac{\left \lfloor \frac n 2 \right \rfloor}{6}$ Last edited by romsek; October 3rd, 2016 at 06:42 PM. 

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