Question a)
Let $n$ $\ge $ $2$ be the number of students who are writing an exam. Each of these students has a uniformly random birthday, which is independent of the birthdays of the other students. We ignore leap years; thus, the year has $365$ days. Define the event:
A = "at least two students have their birthday on December 1"
What is $Pr(A)$ in terms of n.
Attempt:
I am tempted to use the binomial distribution for this to get:
$n\choose{2}$ $.$ $(\frac{1}{356}$)$^{2}$ $.$ $(\frac{364}{365})$$^{n-2}$
However, if I use n=2, I am getting the $Pr(A)$ to be $2.24$ $ * $10$^{-5}$
If I attempt $1-Pr(A)$ = $1 - $ $n\choose{2}$ $.$ $(\frac{1}{356}$)$^{2}$ $.$ $(\frac{364}{365})$$^{n-2}$ = $0.999978$ which just doesn't seem feasible for this case.
Don't think these are the right though.
Question b)
Let $X$ $=$ $(1,2,3,...100$). We choose a uniformly random subset $Y$ of $X$ having size $17$. Define the event:
A = "$4$ $\varepsilon$ $Y$ or $7$ $\varepsilon$ $Y$"
What is $Pr(A)$?.
Answer: $0.285050$
Attempt:
In this I determined the sample space to be $|S|$ = $100\choose7$
For $|A|$ / $|S|$ = {$100\choose1$ + $100\choose1$ - $100\choose2$} / {$100\choose7$}
Not getting the right answer. My thinking is you have to choose 1 element from 100 or again choose 1 element from 100 and using inclusion/exclusion you subtract the case both were chosen.
Question c)
Consider a uniformly random permutation of the set $(1,2,3,...77$). Define the event:
A = "In the permutation, both $8$ and $4$ are to the left of $3$"
What is $Pr(A)$?.
Answer: $\frac{1}{3}$
I am surprised I got this one wrong!
$|S| = 77!$ (total permutations fo the set is our sample size)
|A| = $77\choose3$ $.$ $3$ $.$ $74!$ / $77!$ = $\frac{1}{2}$
From $77$ positions, must choose $3$ positions for $8$ ,$4$, and $3$. And the remaining $74$ positions can be arranged in $74!$ ways. Where did I go wrong for this?
(a) In your attempt, by using $(\frac{364}{365})^{n-2}$, you imply that nobody else except the 2 students you've chosen were born on 12/1. But the question states "at least two", so you'll have to consider when 3 or more students were born on 12/1.
(b) Use $1-$Pr(Neither 4 or 7 $\in Y$)
(c) 3 is always on the rightmost, so you only choose the order of 8 and 4.