Uniform Probabilty

Oct 2016
2
0
New Jersey
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.
 

romsek

Math Team
Sep 2015
2,679
1,481
USA
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 $n-1$ batteries to make $\dfrac {n-1}{2}$ packs.

In this case $E[t] = \dfrac{n-1}{3}$

So our final expectation is

$E[t] = \dfrac{\left \lfloor \frac n 2 \right \rfloor}{3}$
 

romsek

Math Team
Sep 2015
2,679
1,481
USA
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.

$Pr[\text{life of pack}<t] =1-Pr[\text{life of pack}>t]^2$

In other words in order for the pack to have failed by time $t$ both batteries cannot have lasted longer than $t$

$Pr[\text{life of pack}>1] = 1-t$

$Pr[\text{life of pack}<t]=1-(1-t)^2 = 2t-t^2$

Thus the pdf of the pack lifetime is

$p(t) = \dfrac{d}{dt} 2t-t^2 = 2(1-t)$

The expected lifetime of a pack is thus

$E[t] = \displaystyle{\int_0^1}2t(1-t) = \dfrac 1 3$

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

$E[t | n \text{ batteries }]=\dfrac{\left \lfloor \frac n 2 \right \rfloor}{6}$
 
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