Distributing 12 indistinguishable calls in 7 days

Question

I tried approaching the following problem:

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?

I want to know what would the probability be if we were to assume the calls are not distinguishable. When I was working on it, I used the following reasoning and I think I may be wrong but don't know precisely why is my argument incorrect.

I assumed that I was first going to assign one call to each day of the week (since I don't care about the order and the calls are not distinguishable I am assuming that there is only one way to do this).

So I would then be left with $5$ out of the $12$ calls. I used the combination with repetition $\binom{n+k-1}k$, where I am taking $n=7$ (since there are 7 days in a week), and $k=5$ (since I am still left with 5 indistinguishable calls). $\binom{7+5-1}5=462$.

When dividing this number over the total number of possible ways of arranging the 12 calls in 7 days $\binom{7+12-1}{12}$, I get $0.02488$

If I am way off please give me a hint as to what am I doing wrong when following this reasoning. I was looking at a solution for the case when the number of calls are distinguishable, and the answer is a much greater probability ($0.2285$).

Answer 1

For each $1\leq k\leq 7$ let $A_k$ be the event that no phone calls occured on day $k$. You're looking to compute $P\Big((A_1\cup\ldots\cup A_7)^C\Big)$. Note that for any $1\leq k \leq 7$ fixed we have $$P(A_1\cap\ldots\cap A_k)=\Big(\frac{7-k}{7}\Big)^{12}$$ So from inclusion exclusion, $$\begin{eqnarray*}P(A_1 \cup \ldots \cup A_7) &=& \sum_{k=1}^7(-1)^{k-1}{7 \choose k}P(A_1 \cap \ldots \cap A_k) \\ &=&\sum_{k=1}^7(-1)^{k-1}{7 \choose k}\Big(\frac{7-k}{7}\Big)^{12} \end{eqnarray*}$$ Therefore $$ \begin{eqnarray*}P\Big((A_1\cup \ldots \cup A_7)^C\Big)&=&1-P(A_1\cup \ldots \cup A_7) \\ &=& 1-\sum_{k=1}^7(-1)^{k-1}{7 \choose k}\Big(\frac{7-k}{7}\Big)^{12} \\ &\approx& 0.2285\end{eqnarray*}$$ With this interpretation we're treating the phone calls as distinguishable. Moreover, thanks to @lulu, I now know (and will never forget) that the patterns counted by stars and bars are not equiprobable. You need to proceed with extreme caution when using stars and bars to evaluate probabilities. For example, consider distributing $3$ indistinguishable balls in $2$ distinguishable bins. The probability that each bin contains at least one ball equals $1/2$. The probability all balls are distributed in the first bin is $1/4$, and the same for the second bin.

Answer 2

When the callings are indistinguishable your solution is correct and nice ! As far as i have understood you want to find the probability when the callings are distinguishable and each day has at least $1$ calling.

In these type of question , when it is said that at least one object must be in distinguishable boxes , we firstly assume that boxes are indistinguishable . Then, the question turned out to be distinguishable objects into indistinguishable boxes. So , you must remember stirling numbers of second type. Hence , we should find $\color{blue}{S}(12,7)=627396$.

However, we know that the boxes ,i.e, days were different , so we should multiply the foregoing result by $7!$.

Thus , the number of distributing $12$ different calling into $7$ different days where each day has at least $1$ calling is $627396 \times 7!$

Now we should find the number of distributing $12$ distinguishable calling in $7$ distinguishable day without restriction. When we solve these types of question , we prefer exponential generating functions.So , you can do it by finding the coefficient of $\frac{x^{12}}{12!}$. We can find it by $7^{12}$ because we deal with $e^{7x}$

As a result probability = $\frac{627396 \times 7!}{7^{12}}=0,2284524404...$

You can find more information in :How many ways are there to distribute $85$ different candies to $30$ children if each children has at least $2$ candies