# Discrete Probability: Uniformly random subsets, permutations and birthday probability

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?

• You must multiply with 2 not 3 at c). You have 2 possibilites for 8 and 4, 3 is always last one on thos 3 position you choose – Aqua Dec 1 '18 at 22:43

(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 $$\in Y$$)