The question is split into two parts.
- Determine an expression for the probability that at least one individual selects their own mobile phone. Let $L(n)$ denote this expression.
- 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.