I have some questions about the solution to this problem for $n=3$. The problem goes:
"Suppose that each of three men at a party throws his hat into the center of the room. The hats are first mixed up and then each man randomly selects a hat. What is the probability that none of the three men selects his own hat?"
One of the solutions reads Equation 1
$P($no man selects his hat$) = 1 - P($at least one man selects his own hat$) = 1 - P(E1∪E2∪E3)$
I am having a hard time understanding why
$P($at least one man selects his own hat$) = P(E1∪E2∪E3)$
I can understand the complement, basically if $E_i$ is the event that man $i$ picks his own hat, then we are looking for the complement of $\cap E_i$ which equals $\cup E_i^c$. However is the union of events to be used when we answer the question of at least one event occurring?
My second question is this:
while I was trying to solve this problem for $n=3$, I used the following reasoning:
$P($no man selects his hat $) = 1 - P(E_1) - P(E_1E_2) - P(E_1E_2E_3)$
where $E_1$ is the event that only one man picks his own hat $E_1E_2$ is the event that two men pick their own hat, and $E_1E_2E_3$ the event that all did.
In my computations, I ended up with $1/3$, which is the same probability we get if we use $ \ $ Equation 1. Is my reasoning correct?
You have to think about what kind of outcomes are possible when we say "at least one man selects his own hat." If $E_i$ represents the outcome that man $i$ selects his own hat, then $E_1 \cup E_2 \cup E_3$ includes the outcome where all three select their own hat; it also includes the outcome where exactly one man selects his own hat. Note it is impossible for exactly two men to select their own hat--since if this happens, the remaining hat is necessarily selected by its owner. So it just so happens in this case that $$\begin{align*} \Pr[E_1 \cup E_2 \cup E_3] &= \Pr[E_1 \cap \bar E_2 \cap \bar E_3] + \Pr[\bar E_1 \cap E_2 \cap \bar E_3] + \Pr[\bar E_1 \cap \bar E_2 \cap E_3] \\ &\quad + \Pr[E_1 \cap E_2 \cap E_3]. \end{align*}$$ I recommend that you draw a Venn diagram to understand this.
In regard to your second question, you are being sloppy with your notation, and as a result, you are not enumerating all of the relevant outcomes. Your calculation worked only because of coincidence. A correct calculation requires you to consider the outcomes of all three men's selections, so if you write $\Pr[E_1]$, this is just the unconditional probability of the first man selecting his own hat, which is $1/3$. But this probability also counts the event where all three men select their own hat, since $$E_1 \cap E_2 \cap E_3 \subset E_1.$$
As an exercise, try to perform the same computation for four men.