Solution Set 2: Statistics
No. 1
No, it doesn’t mean that the coin toss is fair because $p$ is undetermined. You would have
\[\text{Bern}(1/2,1/2)\]for a fair coin toss.
Since we are talking about coins, the smallest value $p$ could have is near 0. $p$ can’t be 0 otherwise we don’t really have a coin. Coins have 2 sides.
No. 2
Even if $\bar{X}$ is positive, the values in the sample can still be negative. On average, we have positive values. Take
\[\frac{-1+2+2}{3} > 0\]But still had negative values in the sample.
No. 3
Using the pythagorean theorem we see
\[A^2 + B^2 = c^2\]For random $A$ and $B$. Now, since one side is written in terms of the other two, it can be random, but not unconditional on the values of the other two. So, this side, $c$ doesn’t really vary independently of the other two; once we know $A$ and $B$, we know $c$ for sure. There are 2 random variables. The third one is $c|A,B$ by construction, which doesn’t vary at all. $c$ does, though.
So, at most 2 vary independently.
No. 4
They will have it the same day again regardless of what day it was! This always happens. Wait 7 years from that day, assuming no leap years. The chance is 1. Assuming leap years we would have to wait less but it still happens.
No. 5
The chance is 0 since 13 is a prime number. We can’t factor it into a product using numbers 1-6. Charlie will eat the sandwich.