# Probability of at least two people having a specific birthday and a subset containing two elements

1. Let $n ≥ 2$ be the number of people in a room. Each of these people have a uniformly random and independent birthday. The year has $365$ days (ignore leap years).

Define the event $A =$ “at least $2$ people have their birthday on December $14$”.

What is Pr(A)?

Answer: $1-(\frac{364}{365})^n -n \cdot \frac{1}{365} \cdot (\frac{364}{365})^{n-1}$

Can someone break down the answer and tell me what each part counts?

$Pr($at least 2$) = 1 - Pr($at least 1$) - Pr($ no birthdays on Dec. 14$)$

$Pr($at least 1 birthday on December 14$)$:
I think that $(\frac{364}{365})^n$ represents that there are $n$ people who have a choice of $364$ out of $365$ birthdays (excluding December 14). Since there are $n$ people and they all have a choice of $364$, then: $(\frac{364}{365})^n$

This is where I get confused:

$Pr($no birthdays on Dec. 14$)$
How does $n \cdot \frac{1}{365} \cdot (\frac{364}{365})^{n-1}$ count the people who do do not have a birthday on December 14th?

2. Let $X = \{1,2,..., n\}.$ We choose a uniformly random subset $Y$ of $X$ having size $17$.
Define the event:

$A = 4 \in Y$ or $7 \in Y$

What is Pr(A)?

Answer: $\frac{2 \cdot \binom{99}{16} - \binom{98}{15}}{\binom{100}{17}}$

Can someone break this down as well? Where is the numerator coming from?

Thank you.

## 1 Answer

$\frac{364}{365}$ is the probability that a single person does not have their birthday on December 14. The probability that none of the $n$ people have their birthday on that date is thus $\left(\frac{364}{365}\right)^n$. (By multiplication of independent probabilities.)

The probability that a particular person (Alice, say) has their birthday on December 14 is $\frac{1}{365}$; the probability that none of the $n-1$ others has their birthday on that date is $\left(\frac{364}{365}\right)^{n-1}$. The probability of those things happening together is their product: $\frac{1}{365}\left(\frac{364}{365}\right)^{n-1}$. We now multiply this by the probability that $n$ to account for the fact that any of the $n$ people might take Alice's place (including Alice herself, of course), to yield $n\cdot\frac{1}{365}\left(\frac{364}{365}\right)^{n-1}$.

In problem $2$, presumably $n = 100$. The denominator is the total number of ways to pick a subset of size $17$: $\binom{100}{17}$. The numerator should then be the number of ways to pick a subset that contains either $4$ or $7$. Consider the number of ways to pick a subset that contains the number $4$. It is the number of ways to pick the number $4$ (which is just $1$) times the number of ways to pick the other $16$ numbers in the subset out of the remaining $99$: $\binom{99}{16}$.

That just accounts for the $4$. The $7$, by the same line of reasoning, also gives us $\binom{99}{16}$ ways. So we can just add them together!

There's one last step. We must account for the fact that this numerator currently doubly counts subsets that contain both $4$ and $7$, so we need to subtract that number of subsets. This is the number of ways to choose $4$ (which is $1$), times the number of ways to choose $7$ (which is $1$), times the number of ways to choose the other $15$ numbers out of the remaining $98$: $\binom{98}{15}$. So the final numerator is

$$2\binom{99}{16} - \binom{98}{15}$$

This is an example of a general principle called inclusion-exclusion (q.v.).