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 $50\choose 4$ 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) $P(A\cup B) = P(A) + P(B)-P(A\cap B$).
2) $P(A|B) = \frac{P(A\cap B)}{ P(B)}$
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 $P(A)$
$1 = P(A) + P(A|B) / P(A\cap B) – P(A\cap B)$
$1 = P(A) + 2 [P(A|B) / 4 – 1]$
This was just not making sense. Don't know how else to get the right equality with the given conditions
Best Answer
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 ${50\choose 4}$ 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 $$P = {{50\choose 4} \cdot 4 \cdot 46!\over 50!} = {1\over 6} $$
b) Since $Pr(A|B) = Pr(B|A)$ we have $${P(A\cap B)\over P(B)} = {P(A\cap B)\over P(A)} \implies P(A)= P(B)$$
Now acording to PIE for probability we have $$1= P(A\cup B) = P(A)+P(B) -P(A\cap B) < 2P(A) - 0$$