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The question is split into two parts.

  1. Determine an expression for the probability that at least one individual selects their own mobile phone. Let $L(n)$ denote this expression.
  2. Determine the exact value of $lim_{n\to\infty}L(n)$.

I did the first part as follows:

The sample space is $|\Omega|=n!$, the number of different ways the phones can be distributed among the participants. Let $A$ be the event that no person selects their phone and $A_i$ the event that person $i$ doesn't select their phone. Then $$P(A)=P(A_1\cap\cdots\cap A_n)$$$P(A^c)$ would be the probability that at least one person selects their phone and $P(A_i^c)$ the probability that person $i$ does select their phone with $|A_i^c|=(n-1)!$. Then $P(A_i^c)=\frac{(n-1)!}{n!}$ and $P(A_{i_1}^c\cap\cdots\cap A_{i_k}^c)=\frac{(n-k)!}{n!}$ and via de Morgan and inclusion exclusion $$P(A^c)=P(A_1^c\cup\cdots\cup A_n^c)=\sum_{k=1}^n(-1)^{k-1}\sum_{1\leq i_1<\cdots<i_k\leq n}P(A_{i_1}^c\cap\cdots\cap A_{i_k}^c)$$ For the inclusion exclusion we use the fact that $$\sum_{1\leq i_1<\cdots<i_k<n}P(A_{i_1}^c\cap\cdots\cap A_{i_k}^c)={n\choose k}P(A_{i_1}^c\cap\cdots\cap A_{i_k}^c)$$ Substituting all of this into the inclusion exclusion formula yields \begin{align*}P(A^c)&=\sum_{k=1}^n(-1)^{k-1}\frac{(n-k)!}{n!}{n\choose k}\\&=\sum_{k=1}^n(-1)^{k-1}\frac{(n-k)!}{n!}\frac{n!}{(n-k)!k!}\\&=\sum_{k=1}^n(-1)^{k-1}\frac{1}{k!}\\&=L(n)\end{align*} As with any probability question without answeres, I am not very confident in my answer. This is especially so considering that the second question asks for the exact value of an infinite series containing $(-1)^{k-1}$. I know that it will most likely converge to some value but I have no clue how.

Nicojwn
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  • See Derangements. For what it's worth, I may be mistaken, but I suspect that recursion provides an alternative approach to the problem. – user2661923 May 09 '22 at 16:34
  • With regards to the value of an infinite series... recall that $\text{exp}(x)$ is defined as $1+\sum\limits_{k=1}^\infty \frac{x^k}{k!}$ (equivalently, having started from zero assuming you are comfortable with $0^0=1$). You are expected to be familiar with the number $e$ at this point. – JMoravitz May 09 '22 at 16:39
  • @user2661923 Thank you that helped – Nicojwn May 09 '22 at 16:43
  • @JMoravitz I am very much familiar with $e$ :) and thanks for the reminder – Nicojwn May 09 '22 at 16:44

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