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?
Best Answer
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.