Solution: 2018 Fall Final - 15

Author: Michiel Smid

Question

Consider a uniformly random permutation $a_1,a_2,a_3,a_4$ of the set $\{1,2,3,4\}$. Define the events

A = "$a_1 > a_2$",
B = "$a_4 > a_3$".

Which of the following is correct?

(a)

None of the above.

(b)

All of the above.

(c)

The events $A$ and $B$ are not independent.

(d)

The events $A$ and $B$ are independent.

Solution

I have a gut feeling that they’re independent, but let’s check

Let S be all possible permutations of the set ${1, 2, 3, 4}$: $ |S| = 4! = 24 $
Let's determine A
We can view it untuitively or mathematically
- Intuitively: When $a_1$ and $a_2$ are random values in a set, there's a $ \frac{1}{2} $ chance of $a_1 > a_2 $
  $ Pr(A) = \frac{1}{2} $
- Mathematically: Start off by choosing 2 random values from the set ${1, 2, 3, 4}$ to take the place of $a_1$ and $a_2$: $ \binom{4}{2} $
  There's only 1 correct permutation of the 2 values that allow $ a_1 > a_2 $: $ 1 $
  The value at position $a_3$ can be one of the 2 remaining values: $ 2 $
  The value at position $a_4$ can be the remaining value: $ 1 $
  $ |A| = \binom{4}{2} \cdot 1 \cdot 2 \cdot 1 $
  $ |A| = 12 $
  $ Pr(A) = \frac{12}{24} $
  $ Pr(A) = \frac{1}{2} $
Let's determine B
We can view it untuitively or mathematically
- Intuitively: When $a_3$ and $a_4$ are random values in a set, there's a $ \frac{1}{2} $ chance of $a_3 > a_4 $
  $ Pr(B) = \frac{1}{2} $
- Mathematically: Start off by choosing 2 random values from the set ${1, 2, 3, 4}$ to take the place of $a_3$ and $a_4$: $ \binom{4}{2} $
  There's only 1 correct permutation of the 2 values that allow $ a_3 > a_4 $: $ 1 $
  The value at position $a_1$ can be one of the 2 remaining values: $ 2 $
  The value at position $a_2$ can be the remaining value: $ 1 $
  $ |B| = \binom{4}{2} \cdot 1 \cdot 2 \cdot 1 $
  $ |B| = 12 $
  $ Pr(B) = \frac{12}{24} $
  $ Pr(B) = \frac{1}{2} $
Let's determine $ A \cap B $
We take any 2 random values from the set ${1, 2, 3, 4}$ to take the place of $a_1$ and $a_2$: $ \binom{4}{2} $
Now, there's only 1 correct permutation of the 2 values that allow $ a_1 > a_2 $: $ 1 $
Smilarly, we take 2 random values from the remaining set: $ \binom{2}{2} $
There's only 1 correct permutation of the remaining 2 values that allow $ a_4 > a_3 $: $ 1 $
$ |A \cap B| = \binom{4}{2} \cdot 1 \cdot \binom{2}{2} \cdot 1 $
$ |A \cap B| = 6 $
$ Pr(A \cap B) = \frac{6}{24} $
$ Pr(A \cap B) = \frac{1}{4} $

Now, let’s check whether they’re independent

$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $

$ \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} $

$ \frac{1}{4} = \frac{1}{4} $

They are independent