I ran across the following problem in Casella and Berger's Statistical Inference (Q1.20, 2nd ed):
My telephone rings 12 times each week, the calls being randomly distributed among the 7 days. What is the probability that I get at least one call each day? (Answer: .2285)
This seems to be equivalent to putting 12 balls into 7 boxes so that there is at least 1 ball in each box. In that case, this should be a fairly straightforward selection-with-repetition problem: since we have at least 1 ball in each box, that means we must actually count the number of ways to put 5 balls into 7 boxes, and divide by the number of ways to put 12 balls into 7 boxes, which would give [(11 choose 6)/(18 choose 6)] or about 0.0249.
I see an answer given here: Statistical Inference Question which approaches it from the bottom up rather than the top down. This method seems reasonable, and I've verified that it gives the authors' desired answer, but what's different about my approach? In both cases, the calls are unordered and identical, the days are ordered and distinct, and repetition is allowed.
Best Answer
Inclusion/exclusion works well.
Take $A_i$ as the event that at least one call is received on day $i$. You're looking to compute the probability of $A_1\cap \dots \cap A_7$. Then $$\begin{eqnarray*}\mathbb{P}\left(A_1 \cap \dots \cap A_7\right) &=& 1- \mathbb{P}\left(\bigcup_{i=1}^7A_i^C\right) \\ &=& 1-\sum_{i=1}^7(-1)^{i-1}{ 7 \choose i}\mathbb{P}\left(A_1^C \cap \dots \cap A_i^C\right) \\ &=& 1-\sum_{i=1}^7(-1)^{i-1}{ 7 \choose i}\left(\frac{7-i}{7}\right)^{12} \\ &\approx & 0.2285\end{eqnarray*}$$