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$150$ balls randomly put into $100$ boxes, each ball could be put into any of these 100 boxes with same probability, after that, on average, how many boxes will be empty? No calculator. Choose one of the following:

A 0-10
B 10-20
C 20-30
D 30-40
E 40-50
F 50-60
G 60-70
I 70-80
J 80-90
K 90-100

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the only thing you can be sure of: there won't be 100 empty boxes... – long tom Mar 23 '13 at 21:25
I mean on average – nkhuyu Mar 23 '13 at 21:29
up vote 1 down vote accepted

Each box gets $1.5$ balls on average and I would guess a Poisson distribution is a pretty good fit. For a Poisson distribution with expected value $\lambda$ the probability of zero is $\exp(-\lambda)$, so we need to calculate $100 \exp(-1.5)$. The square root of $e$ is between $1.6=\sqrt {2.56}$ and $1.7=\sqrt {2.89}$, so $\exp (-1.5) \approx \frac 1{1.6 \cdot 2.7} =\frac 1{4.32}$ This gives $100 \exp (-1.5) \approx 23-24$ and I will choose $20-30$

Checking with a calculator, $100 \exp (-1.5)$ is in fact $22.3$, a little lower than I guessed.

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Great solution, I never thought about Poisson distribution – nkhuyu Mar 23 '13 at 22:05
@nkhuyu: It is based on lots of low probability events. Here the event for a given box is a ball comes in, which is 150 tries with probability 0.01. You can argue that there is some correlation between the boxes, but I think the effect will be quite small. – Ross Millikan Mar 23 '13 at 22:07
FWIW a VB simulation gives approx 22.1 empty per trial... – DJohnM Mar 24 '13 at 3:05

Perhaps you need to calculate somehow the probability of $n$ boxes to be empty $( 0 \leq n \leq 99)$ and then to calculate the expected value.

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Do you consider the boxes to be identical or different? What about the balls? Is this task (if it's a task from some course, book etc.) connected with some particular discrete distribution or just a question? – Igor Mar 23 '13 at 21:43
I don't think you need to calculate the expected value (no calculator allowed, besides calculating $150!$ tends to be huge). I guess the chances for $0$ or $1$ (assuming you calculated those by hand) empty boxes are so overwhelming every sane person would go for option $A$. So I'd put my money on option $A$ – long tom Mar 23 '13 at 21:47

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