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From Statistical Inference Second Edition (George Casella, Roger L. Berger)

"My telephone rings 12 times each week, the calls being randomly distributed among the 7 days. What is the probability that I get a least one call each day?"

The answer is .2285, but I don't know how they got it. My reasoning was as follows:

There are 12 calls and thus 13 places to put "day dividers" to produce possible distributions of calls. There should be 6 day dividers for the week. One possibility:

_1|2_3|4_5_6|7|8||9_10_11_12

(1 on Monday, 2 on Tuesday, 3 on Wednesday, 1 on Thursday, 1 on Friday, 0 on Saturday, 4 on Sunday)

There are 13^6 possible distributions (using this method). In order to satisfy 1 call/day, dividers can't be at the beginning or end, nor can they be on top of one another (signifying a day with no calls). This means there are:

11*10*9*8*7*6

Valid distributions and a probability of: 0.069. Where am I going wrong?

(P.S. This isn't homework as I'm not in school.)

Edit: I don't think each "distribution" is equally likely. That's probably my error. But I still don't know how to get to the correct answer :)

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Does your way of counting really coincide with a uniform distribution of calls over days? (at least that is a natural interpretation of the ill-posed problem) –  Raphael Dec 6 '10 at 10:26
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You can for sure model this using a multinomial distribution but that does not immediately yield a "nice" solution. –  Raphael Dec 6 '10 at 11:14
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1 Answer

up vote 2 down vote accepted

Here is an easy approach to solve the problem. Consider the tuples $(n_1,n_2,\ldots,n_7)$ such that $n_i$ are positive integers satisfying $n_1 + n_2 + \cdots + n_7 = 12$. Further, assume for the moment that $n_1 \geq n_2 \geq \cdots \geq n_7$. You should find that there are only seven such tuples. Take, for example, the tuple $(4,3,1,1,1,1,1)$. The corresponding probability is, according to the multinomial distribution with parameters $n=12$ and $p_1 = p_2 = \cdots = p_7 = 1/7$, $$ \frac{{12!}}{{4!3!1!1!1!1!1!}}\bigg(\frac{1}{7}\bigg)^{4}\bigg(\frac{1}{7}\bigg)^{3}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1} = \frac{{12!}}{{4!3!}}\bigg(\frac{1}{7}\bigg)^{12}. $$ Now, if we don't assume that $n_1 \geq n_2 \geq \cdots \geq n_7$, then you should multiply the above probability by ${7 \choose 2}2$. Doing the same for the other tuples, you obtain seven probabilities. Their sum is the probability you are looking for. I have done the calculation and the result is $\approx 0.228452440447$.

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Gotcha, this makes sense. Thank you! –  Stefan Mai Dec 6 '10 at 21:19
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