Skip to content

Commit 2c5281b

Browse files
committed
Fixed errors.
1 parent 09abde8 commit 2c5281b

4 files changed

Lines changed: 17 additions & 13 deletions

File tree

notes/courses/MATH-UA-333/02-axioms.md

Lines changed: 9 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -9,24 +9,28 @@ A **sample space** $S$ is the set of all possible outcomes, and an **event** is
99

1010
## Probability measure axioms
1111

12-
A probability measure $P$ assigns numbers to events such that $0\le P(E)\le 1$, $P(\emptyset)=0$, $P(S)=1$, and $P(E_1\cup E_2\cup\cdots)=P(E_1)+P(E_2)+\cdots$ for disjoint events. Probabilities can be interpreted as areas in a Venn diagram.
12+
A probability measure $P$ assigns numbers to events such that $0\le P(E)\le 1$, $P(\emptyset)=0$, $P(S)=1$, and $P(E_1\cup E_2\cup\cdots)=P(E_1)+P(E_2)+\cdots$ for disjoint events. Probabilities can be interpreted as areas in a Venn diagram.
1313

1414
### Identities and the inclusion–exclusion principle
1515

1616
From additivity one derives useful identities:
1717

1818
- Complement: $P(A^c)=1-P(A)$.
1919
- Decomposition: $P(B)=P(A\cap B)+P(B\setminus A)$.
20-
- Union: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, and more generally for $n$ events the inclusion–exclusion formula sums intersections of increasing size with alternating signs. The union bound states $P(E_1\cup\cdots\cup E_n)\le P(E_1)+\cdots+P(E_n)$.
20+
- Union: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, and more generally for $n$ events the inclusion–exclusion formula sums intersections of increasing size with alternating signs. The union bound states $P(E_1\cup\cdots\cup E_n)\le P(E_1)+\cdots+P(E_n)$.
2121

2222
## Uniform probability on finite spaces
2323

24-
If $S$ has finitely many equally likely outcomes, then $P(E)=\frac{\#E}{\#S}$.
25-
> Example: two dice have $36$ equally likely outcomes; the event “sum = 7” contains six outcomes, so the probability is $6/36=1/6$. Drawing 7 cards from a 52‑card deck, the probability of getting a four‑of‑a‑kind is
24+
If $S$ has finitely many equally likely outcomes, then
2625
$$
27-
P(E)=\frac{13\cdot {48\choose 3}}{{52\choose 7}}.
26+
P(E)=\frac{\text{Number of elements in } E}{\text{Number of elements in } S}.
2827
$$
2928

29+
> Example: two dice have $36$ equally likely outcomes; the event “sum = 7” contains six outcomes, so the probability is $6/36=1/6$. Drawing 7 cards from a 52‑card deck, the probability of getting a four‑of‑a‑kind is
30+
> $$
31+
> P(E)=\frac{13\cdot {48\choose 3}}{{52\choose 7}}.
32+
> $$
33+
3034
In the birthday problem with 23 people, the probability that at least two share a birthday is $1-\frac{365\cdot364\cdots(365-22)}{365^{23}}$, which exceeds $\tfrac{1}{2}$.
3135

3236
## Examples and applications

notes/courses/MATH-UA-333/04-05-drv.md

Lines changed: 6 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -16,7 +16,7 @@ $$
1616
> Examples:
1717
> - Fair die $X\in\{1,\dots,6\}$: $\E[X]=3.5$.
1818
> - Heads in three fair flips $Y\in\{0,1,2,3\}$: $\E[Y]=1.5$.
19-
> - Indicator $\mathbf{1}_A$: $\E[\mathbf{1}_A]=P(A)$.
19+
> - Indicator ${\mathbf{1}}_A$: $\E[{\mathbf{1}}_A]=P(A)$.
2020
2121
## Linearity of expectation
2222
For random variables $X,Y$ and scalar $a\in\mathbb{R}$,
@@ -26,7 +26,7 @@ $$
2626
No independence required.
2727

2828
**Counting-by-indicators trick**
29-
If $X=\sum_{i=1}^n \mathbf{1}_{A_i}$ counts how many events $A_i$ occur, then
29+
If $X=\sum_{i=1}^n {\mathbf{1}}_{A_i}$ counts how many events $A_i$ occur, then
3030
$$
3131
\E[X] = \sum_{i=1}^n P(A_i).
3232
$$
@@ -39,15 +39,15 @@ $$
3939
$$
4040
Useful identities:
4141
- For constants $a,b$: $\operatorname{Var}(aX+b)=a^2 \operatorname{Var}(X)$.
42-
- For an indicator: $\operatorname{Var}(\mathbf{1}_A)=P(A)\big(1-P(A)\big)$.
43-
- If $X=\sum_{i=1}^n \mathbf{1}_{A_i}$ with independent indicators, then
42+
- For an indicator: $\operatorname{Var}({\mathbf{1}}_A)=P(A)\big(1-P(A)\big)$.
43+
- If $X=\sum_{i=1}^n {\mathbf{1}}_{A_i}$ with independent indicators, then
4444
$$
4545
\operatorname{Var}(X)=\sum_{i=1}^n P(A_i)\big(1-P(A_i)\big).
4646
$$
4747

4848
## Inclusion–Exclusion via indicators
49-
Using $\mathbf{1}_{A^c} = 1 - \mathbf{1}_A$ and expanding
50-
$\mathbf{1}_{\cup_i E_i} = 1 - \prod_{i=1}^n (1-\mathbf{1}_{E_i})$, then taking expectations yields
49+
Using ${\mathbf{1}}_{A^c} = 1 - {\mathbf{1}}_A$ and expanding
50+
${\mathbf{1}}_{\cup_i E_i} = 1 - \prod_{i=1}^n (1-{\mathbf{1}}_{E_i})$, then taking expectations yields
5151
$$
5252
P\Big( \bigcup_{i=1}^n\ E_i \Big)
5353
= \sum_{k=1}^n (-1)^{k-1}

notes/courses/MATH-UA-333/index.json

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -1,6 +1,6 @@
11
[
22
{
3-
"slug": "01-intro",
3+
"slug": "01-combinatorics",
44
"title": "1 - Basic combinatorics",
55
"date": "2025-09-04"
66
}

notes/js/course.js

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -41,5 +41,5 @@ fetch(`/notes/courses/${slug}/index.json`)
4141
})
4242
.catch(e => {
4343
console.error(e);
44-
listEl.innerHTML = "<p>Failed to load notes.</p>";
44+
listEl.innerHTML = "<p>Notes are unavailable at the moment, they might not be uploaded yet.</p>";
4545
});

0 commit comments

Comments
 (0)