# expected value of battery drawing problem

This is the question:

In a box, George has $m$ batteries of which $n$ are dead. He tests them randomly and one by one. Every time that a good battery is drawn, he will return it to the box; every time the dead battery is drawn, he will replace it by a good one.

(i) Determine the expected value of number of good batteries in the box after $n$ of them are checked.

(ii) Determine the probability that on the $n$th draw George draws a good battery.

Will anyone please give me an idea on how to begin solving this problem?

The question sequence seems to be awkward to me. Would I calculate the probability of getting a good battery on nth draw before trying to figure out the expected value?

Even so, I seem to have trouble formulating probability of drawing a good battery on the $n$th draw.

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Questions (i) and (ii) should be solved in this order.

(i) Let $\color{red}{D_k}$ denote the number of dead batteries after $k$ draws. Each of the $n$ batteries which were dead before the draws began has the same chance of being still dead after $k$ draws, hence $\color{red}{\mathrm E(D_k)=np_k}$, where $\color{red}{p_k}$ is the probability that one given (dead) battery was not drawn yet. This corresponds to drawing $k$ times any one of the other $m-1$ (dead or good) batteries hence $$\color{red}{p_k=\left(\frac{m-1}m\right)^k}.$$ (ii) Conditionally on $D_k=i$, the probability to draw a good battery after $k$ draws is $\frac{m-i}m$ hence the non conditional probability $\color{red}{g_k}$ to draw a good battery after $k$ draws is $$\color{red}{g_k=\mathrm E\left(\frac{m-D_k}m\right)=1-\frac{n}mp_k}.$$ In particular, $$\mathrm E(D_n)=n\left(\frac{m-1}m\right)^n\quad\mbox{and}\quad g_n=1-\frac{n}m\left(\frac{m-1}m\right)^n.$$ This is a nice example of the apparent paradox that computing expectations of random variables is in many situations easier than computing probabilities of events.

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For (i), if you look at Wikipedia on the coupon collector's problem you might get an idea. For (ii) if you know the expected number of good ones when $n-1$ have been checked, you can calculate the chance you get a good one next.

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so i'll have to use the markov inequality? – adsisco Nov 9 '11 at 5:40

You should take a look at the negative binomial distribution.

Basically, think about the process as you search for a dead battery. You keep checking for batteries until you find a dead one. Thus, you can treat the 'got a good battery' as a success and 'got a bad battery' as a failure.

Therefore, you are looking at a negative binomial distribution with parameters $p=\frac{m-n}{m}$ and 1 failure (i.e., $r=1$). Here, I am using the wiki's notation for the negative binomial.

To answer qn 1 you have to calculate:

Prob(1 success|p,r) * (m+1) + Prob(2 successes|p,r) * (m+2) + ....+ Prob(n successes|p,r) * (m+n)

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