Question:
a) Let $n$ $\ge$ $3$ be an integer. Consider a uniformly random permutation $a_1$,$a_2$…$a_n$ of the set ${(1,2,…,n})$. Define the events:
A = "$a_n$ = n"
B = "$a_2$ $\gt$ $a_1$"
Are these independent or dependent? Show why?
Attempt:
I have to show $P$($A$$\cap$$B$) = $P(A)P(B)$ to prove independence
If I take $n=3$: I get the set $({1,2,3})$ that can be arranged in $3$ ways that satisfy the
condition of "$a_2$ $\gt$ $a_1$". This is out of the total $8$ choices.
So, $Pr(B)$ $=$ $\frac{3}{8}$
For $Pr(A)$, I don't know what they mean by this. "$a_n$ = n". Isn't $a_n$ always going to be $n$? So, $Pr(A)$ = 1?
For $P$($A$$\cap$$B$) = $\frac{3}{8}$ $.$ 1 = $\frac{3}{8}$
$P$($A$$\cap$$B$) = $P(A)P(B)$ = $\frac{3}{8}$ = Independent!
Question:
b) Let $n$ $\ge$ $5$ be an integer. Consider a uniformly random permutation $a_1$,$a_2$…$a_n$ of the set ${(1,2,…,n})$. Define the events:
A = "$a_1$ = 1"
B = "$a_n$ $=$ $5$"
What is $Pr$($A$$\cup$$B$)?
a) $\frac{1}{n}$ $-$ $\frac{1}{n(n-1)}$
b) $\frac{2}{n}$ $-$ $\frac{1}{n(n-1)}$ (Answer)
c) $\frac{2}{n}$ $-$ $\frac{1}{n^2}$
Attempt:
For $n=5$, I can find the $P(A)$, $P(B)$ to be $\frac{1}{5}$
I have to show $P$($A$$\cup$$B$) = $P(A) $ + $ P(B)$ $-$ $P$($A$$\cap$$B$).
I am struggling to determine $P$($A$$\cap$$B$) for this. Would it just be $\frac{1}{5}$ $.$ $\frac{1}{5}$? That doesn't give me the correct solution for $n=5$ which should be 0.35.
Best Answer
(a)
If you have $3$ numbers and you permute them, you have only $3!=6$ possibilities.
$$Pr(B)=\frac{1}{2}$$ from symmetry.
$$Pr(A)=\frac1{n!}$$
since there is only one sequence that satisfies that condition.
Also $$Pr(A \cap B)=Pr(A) \ne Pr(A)Pr(B)$$
hence they are not independent.
(b)
$$Pr(A)=Pr(B)=\frac{(n-1)!}{n!}=\frac1n$$
$$Pr(A\cap B)= \frac{(n-2)!}{n!}=\frac1{n(n-1)}$$
The idea is while two positions are fixed, we are free to permute the rest.
Try to avoid the temptation of thinking that $P(A\cap B)=P(A)P(B)$ unless you can justify that they are independent.