Solution: 2022 Winter Final - 15

Author: Michiel Smid

Question

Let $\pi_1,\ldots,\pi_{20}$ be a random permutation of $\{1,\ldots,20\}$. Define the events: $ A = \pi_{10}>\pi_{11} $ and $ B = \pi_{11} \gt \pi_{12} $ Which of the following is true?

(a)

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

(b)

None of the above

(c)

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

(d)

$\Pr(A\mid B)$ is undefined

(e)

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

Solution

Warning: I don’t know but I can try

Let S be the set of all possible permutations
$ |S| = 20! $
Let's determine $A$
We could pick 1 and any of the 19 larger integers and the other 18 numbers have 18! posible permutations
We could pick 2 and any of the 18 larger integers and the other 18 numbers have 18! posible permutations
\ldots
We could pick 19 and 20 and the other 18 numbers have 18! posible permutations
$ |A| = 19 \cdot 18! + 18 \cdot 18! + \ldots + 1 \cdot 18! $
$ |A| = 18! \cdot \frac{(19)(20)}{2}$
$ |A| = 18! \cdot (19)(10) $
$ |A| = 18! \cdot 190 $ $ |A| = 10 \cdot 19! $ $ \Pr(A) = \frac{|A|}{|S|} $
$ \Pr(A) = \frac{10 \cdot 19!}{20!} $
$ \Pr(A) = \frac{10}{20} $
$ \Pr(A) = \frac{1}{2} $
Let's determine $\Pr(B)$
We could pick 1 and any of the 19 larger integers and the other 18 numbers have 18! posible permutations
We could pick 2 and any of the 18 larger integers and the other 18 numbers have 18! posible permutations
\ldots
We could pick 19 and 20 and the other 18 numbers have 18! posible permutations
$ |A| = 19 \cdot 18! + 18 \cdot 18! + \ldots + 1 \cdot 18! $
$ |A| = 18! \cdot \frac{(19)(20)}{2}$
$ |A| = 18! \cdot (19)(10) $
$ |A| = 18! \cdot 190 $ $ |A| = 10 \cdot 19! $ $ \Pr(A) = \frac{|A|}{|S|} $
$ \Pr(A) = \frac{10 \cdot 19!}{20!} $
$ \Pr(A) = \frac{10}{20} $
$ \Pr(A) = \frac{1}{2} $
Let's determine $\Pr(A \cap B)$
Well, let's go I guess!
- $\pi_{12}$ could be 1 and $\pi_{11}$ would have to be any of the 19 values smaller than it and $\pi_{10}$ would have to be any value larger than $\pi_{11}$. The remaining 17 values have 17! permutations
- $\pi_{12}$ could be 2 and $\pi_{11}$ would have to be any of the 18 values smaller than it and $\pi_{10}$ would have to be any value larger than $\pi_{11}$. The remaining 17 values have 17! permutations
- \ldots
- $\pi_{12}$ could be 19 and $\pi_{11}$ would have to be 20 and $\pi_{10}$ would have to be any value larger than $\pi_{11}$. The remaining 17 values have 17! permutations
Idk if there's a way to express it but we can try
$ \sum_{i=1}^{18} ( \sum_{j=i+1}^{19} (\sum_{k=j+1}^{20} 1 \cdot 17!) ) $
$ 17! \cdot \sum_{i=1}^{18} ( \sum_{j=i+1}^{19} (\sum_{k=j+1}^{20} 1) ) $
$ 17! \cdot \sum_{i=1}^{20} ( \sum_{j=1}^{i-1} (\sum_{k=1}^{j-1} 1) ) $
$ 17! \cdot \sum_{i=1}^{20} ( \sum_{j=1}^{i-1} ( j - 1 ) ) $
$ 17! \cdot \sum_{i=1}^{20} ( \frac{(i-1)(i-2)}{2}) $
$ 17! \cdot \frac{1}{2} \cdot \sum_{i=1}^{20} ( (i-1)(i-2) ) $
$ 17! \cdot \frac{1}{2} \cdot \sum_{i=1}^{20} ( i^2 - 3i + 2 ) $
$ 17! \cdot \frac{1}{2} \cdot ( \sum_{i=1}^{20} i^2 - \sum_{i=1}^{20} 3i + \sum_{i=1}^{20} 2 ) $
$ 17! \cdot \frac{1}{2} \cdot ( \frac{20 \cdot 21 \cdot 41}{6} - 3 \sum_{i=1}^{20} i + 2 \sum_{i=1}^{20} 1 ) $
$ 17! \cdot \frac{1}{2} \cdot ( \frac{20 \cdot 21 \cdot 41}{6} - 3 \cdot \frac{20 \cdot 21}{2} + 2 \cdot 20 ) $
$ 17! \cdot \frac{1}{2} \cdot ( \frac{20 \cdot 7 \cdot 41}{2} - 3 \cdot \frac{10 \cdot 21}{1} + 40 ) $
$ 17! \cdot \frac{1}{2} \cdot ( \frac{10 \cdot 7 \cdot 41}{1} - 3 \cdot 10 \cdot 21 + 40 ) $
$ 17! \cdot \frac{1}{2} \cdot ( 10 \cdot 7 \cdot 41 - 3 \cdot 10 \cdot 21 + 40 ) $
$ 17! \cdot ( 5 \cdot 7 \cdot 41 - 3 \cdot 5 \cdot 21 + 20 ) $
$ 17! \cdot ( 5 \cdot 7 \cdot 41 - 3 \cdot 5 \cdot 21 + 20 ) $
$ 17! \cdot ( 1435 - 315 + 420 ) $
$ 17! \cdot ( 1540 ) $
ngl, the value is actually wrong. Basically the probability of A intersect B is smaller than the product of the 2 probabilities

There’s no way this is the actual way to do the question, right? This is bullshit. I am LITERALLY losing my sanity