Xrfgtjtk

Question:

a) Consider a uniformly random permutation of the set ${(1,2,...,50})$. Define the events:

A = "in the permutation, both $8$ and $4$ are to the left of both $1$ and $2$".

What is $Pr(A)$?

Answer: $frac{1}{6}$

Attempt:

I'm assuming that the $|S|$ is going to be $50!$ based on the permutation formula.

The tricky part is figuring out the event. From 50 positions, 2 positions of 8 and 4 should be fixed in a way that they are to the left of 1 and 2. Would that just be $50choose 4$ ways? I am very confused about how to count these.

Question:

b) Let A and B be two events in some sample space. You are given that:

$Pr(A|B) = Pr(B|A)$

$Pr$($A$$cup$$B$) = $1$

$Pr$($A$$cap$$B$) $ > $ $0$

Which of the follwoing is true?

a) $Pr(A)$ $ < $ $frac{1}{2}$

b) $Pr(A)$ $ > $ $frac{1}{2}$ (Answer)

c) $Pr(A)$ $ < $ $1$

d) $Pr(A)$ $ < $ $0$

Attempt:

I narrowed down the equations I needed to use to these:

1) $P(Acup B) = P(A) + P(B)-P(Acap B$).

2) $P(A|B) = frac{P(Acap B)}{ P(B)}$

I tried using the given condition with these equations to hopefully get the answer but I'm getting nowhere it seems.

Using 1) I tried getting $P(A)$

$1 = P(A) + P(A|B) / P(Acap B) - P(Acap B)$

$1 = P(A) + 2 [P(A|B) / 4 - 1]$

This was just not making sense. Don't know how else to get the right equality with the given conditions

a) For all those 4 numbers first you choose 4 places where are they going to stand in a permutation, that you can do on ${50choose 4}$ ways and then on first two among chosen are 8 and 4 and on the other two are 1 and 2, so we have two posibilites for 8 and 4 and two posibilites for 1 and 2. Thus $$P = {{50choose 4} cdot 4 cdot 46!over 50!} = {1over 6} $$

b) Since $Pr(A|B) = Pr(B|A)$ we have $${P(Acap B)over P(B)} = {P(Acap B)over P(A)} implies P(A)= P(B)$$

Now acording to PIE for probability we have $$1= P(Acup B) = P(A)+P(B) -P(Acap B) < 2P(A) - 0$$

(a): There are $binom{50}{4}$ ways to choose where those four elements go, $(50-4)!$ ways to order the rest of the elements... and we don't care. All that matters is the relative order of those four elements, something that looks the same regardless of that choice. There are really only $4!$ possibilities that make a difference.

(b): We can't find exactly what $P(A)$ is - but we can show that it's the same as $P(B)$, and use that to constrain them.