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Suppose that $10$ balls are put into $5$ boxes, with each ball independently being put in box $i$ with probability $ p_i, \sum_{i=1}^{5} p_i = 1 $

A) Find the expected number of boxes that do not have any balls.

Attempt: Let $X$ denote the number of boxes without balls. This means $$ EX = \sum_{x=0}^{4} x P(X=x) $$ Since we know each ball must go into at least one box, we cannot have 5 empty boxes, so that is why I sum to 4. I then said $P(X=j) = P(j{}\,\text{boxes with no balls})= {5 \choose j}(1-p_i)^{10}$ So $$EX = 0 + \sum_{j=1}^{4} j{5 \choose j}(1-p_i)^{10}$$ Is it ok?

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    No. It makes no sense since it contains a free index $i$.2012-12-06

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The probability that box $i$ remains empty is $(1-p_i)^{10}$. By linearity of expectation, the expected number of boxes that remain empty is $\sum_i(1-p_i)^{10}$.

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    You could maybe add that you are using Bernoulli RV, may not be clear from is point of view2012-12-06
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    Is there a way to get the right answer using my method? Instead of finding the prob that a particular box was empty, I found the prob that $i$ boxes was empty.2012-12-06
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    @CAF: You can certainly do what you describe in this comment, but that's pretty far from what you write in the question. You can't use $\binom5j$ because you'd have to multiply that with some probability that's the same for all selections of $j$ empty boxes, but the boxes all have different probabilities of being empty. So if you want to do it as you describe in the comment, you have to add up all the individual contributions; doing that and then bring that into the simple form $\sum_i(1-p_i)^{10}$ is going to be quite a pain.2012-12-06
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    Ok, thanks. The next question was find the expected number of boxes with 1 ball in it. So the prob that one box has one ball in it is $ {10 \choose 1} (p_i) (1-p_i)^9 $ so then I just sum this from 1 to 5?2012-12-06
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    @CAF: Exactly. Note that if you also do this for $2,\dotsc,10$ balls per box and take the sum of all these expected numbers, you can then swap the two sums and evaluate the inner sum to $1$, and then the outer sum yields $5$ as it must.2012-12-06
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    Are you talking about a double sum here? (the words inner/outer hint this) if so, why? Could you give me an example of what you mean?2012-12-06
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    @CAF: I mean $$ \sum_{j=0}^{10}\sum_{i=1}^5\binom{10}jp_i^j(1-p_i)^{10-j}=\sum_{i=1}^5\sum_{j=0}^{10}\binom{10}jp_i^j(1-p_i)^{10-j}=\sum_{i=1}^5(p_i+1-p_i)^{10}=\sum_{i=1}^51=5 $$2012-12-06
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    So you are essentially summing over all possible values of $j$, $j$ being the number of balls in each box. I am also used to computing expectation by $EX = \sum_{i} x P(X=x) $why do we not use that here? In particular, I am missing that 'x' on the above computations.2012-12-06
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    @CAF: On your first sentence: Yes. On the question: We *are* using that here. If an event occurs with probability $p$ and $X$ is the number of its occurrences, then $EX=\sum_x xP(X=x)=0\cdot(1-p)+1\cdot p=p$. So each of the probabilities we're summing is the expected number of boxes with one particular index that's empty (or has $1$ ball, or whatever), and then by linearity of expectation we just have to sum over the index to get the expected number of boxes with any index.2012-12-06
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    Taking the case where we want the expected number of boxes that have zero balls in them, I understand that the probability that a box is empty is $ (1-p_i)^{10}$ and so we sum this 5 times (for each box). So that is the P(X=x) part. If X is the number of balls in each box, then using what I know for EX, what is x? (The x in x P(X=x))2012-12-06
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    I suppose where I am getting a little confused is that the proposed answer $$\sum_{i} (1-p_i)^{10}$$ is similar to $$P(\cup_{i=1}^{5} E_i ) = P(\sum_i E_i)$$ which is essentially what we have above but the above is an expectation not a probability. (it is similar, but not the same since the E_i are not disjoint). I am still struggling to see why the definition of expectation is being used here. I.e we want xP(X=x) but there is no x premultiplying P(X=x) in our answer( we have just summed P(X=x) for x from 1 to 5, without saying for example 1P(X=1) + 2P(X=2) +...+ 5P(X=5). Can you help? Thanks!2012-12-08
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    @CAF: My previous comment explicitly demonstrated how $EX\sum_xxP(X=x)$ is being used. The factors $0$ and $1$ are the values of $x$ and the factors $(1-p)$ and $p$ are the corresponding values of $P(X=x)$. I don't know how I can make it any more explicit than that.2012-12-09
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    Ok, I understand now. Is $'1'$ and $'0'$ arbritary though? Could I have said, for example, that $X_i = 3,$ if the box $i$ contained 0 balls and $100$ otherwise. If not, what is special about $0$ and $1$? Thanks2012-12-11
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    Could it be that $X$ is an indicator RV?2012-12-11
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    @CAF: You're making things a lot more complicted than they are. $0$ is the number of empty boxes that box $i$ contributes to the total number of empty boxes if it is not empty, and $1$ is the number of empty boxes that box $i$ contributes to the total number of empty boxes if it is empty. That's all there is to it. Yes, this is an indicator random variable for the event that box $i$ is empty, but you don't need such a fancy term to understand what's going on here. The total number of empty boxes is a sum of five contributions for the five boxes, and each contribution is either $0$ or $1$.2012-12-12
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    But why do we use $'0'$ and $'1'$ to represent the case of not empty/ empty?2012-12-12
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    @CAF: I'm sorry, I don't know how to make it any clearer. A non-empty box is $0$ empty boxes and an empty box is $1$ empty box. This is not representing, just a trivial form of counting. Try thinking less complicated, and it will suddenly become very obvious.2012-12-12