moving stuff around
This commit is contained in:
+203
@@ -0,0 +1,203 @@
|
||||
{
|
||||
"cells": [
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "a6732353-51d5-4478-9cf8-5834e57e5a4e",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"# Chapter 1 Notes"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "48d9ec9e-83da-40ca-ae79-3c45f8af137c",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Main Concepts\n",
|
||||
"\n",
|
||||
"Outcome: A result of a random experiment.\n",
|
||||
"\n",
|
||||
"Sample Space: The set of all possible outcomes.\n",
|
||||
"\n",
|
||||
"Event: A subset of the sample space.\n",
|
||||
"\n",
|
||||
"Inclusion-exclusion principle holds for probability\n",
|
||||
"\n",
|
||||
"Consider a sample space S. If S is a countable set, this refers to a discrete probability\n",
|
||||
"mode\n"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "7ac122be-50b2-423c-b88f-e4b3327b21bd",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Example Problems"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "7fb87a35-a470-4d98-935f-80c814e3f95d",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"Example 1.5 - soln\n",
|
||||
"\n",
|
||||
"- there are 10 people with white shirts and 8 people with red shirts;\n",
|
||||
"- 4 people have black shoes and white shirts\n",
|
||||
"- 3 people have black shoes and red shirts\n",
|
||||
"- the total number of people with white or red shirts or black shoes is 21\n",
|
||||
"\n",
|
||||
"Let A be the set of people with white shirts, B be the set of people with red shirts and let C be the set of people with black shoes.\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"|A|=10 \\\\\n",
|
||||
"|B|=8 \\\\\n",
|
||||
"|A \\cap C| = 4 \\\\\n",
|
||||
"|B \\cap C| = 3 \\\\\n",
|
||||
"|A \\cup B \\cup C| = 21\n",
|
||||
"\\end{align*}\n",
|
||||
"\n",
|
||||
"Now we solve for $|C|$:\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"|A| + |B| + |C| - |A \\cap B| - |A \\cap C| - |B \\cap C| + |A \\cap B \\cap C| = 21 \\\\\n",
|
||||
"10 + 8 + |C| - 0 - 4 - 3 - 0 = 21 \\\\\n",
|
||||
"18 + |C| - 7 = 21 \\\\\n",
|
||||
"|C| + 11 = 21 \\\\\n",
|
||||
"|C| = 10\n",
|
||||
"\\end{align*}\n",
|
||||
"\n",
|
||||
"$\\therefore$ number of people with black shoes is 10\n"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "72b40733-531a-48f3-9879-75601684afc2",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"Example 1.11 - soln\n",
|
||||
"\n",
|
||||
"Suppose we have the following information:\n",
|
||||
"1. There is a 60 percent chance that it will rain today.\n",
|
||||
"2. There is a 50 percent chance that it will rain tomorrow.\n",
|
||||
"3. There is a 30 percent chance that it does not rain either day.\n",
|
||||
"\n",
|
||||
"T = rains\n",
|
||||
"F = no rain\n",
|
||||
"\n",
|
||||
"$S = \\{(F, F), (F, T), (T, F), (T, T)\\}$\n",
|
||||
"\n",
|
||||
"$P((T, F) \\cup (T, T)) = 0.6$\n",
|
||||
"\n",
|
||||
"$P((F, T) \\cup (T, T)) = 0.5$\n",
|
||||
"\n",
|
||||
"$P((F, F)) = 0.3$\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"P(S) = 1 \\\\\n",
|
||||
"P(\\{(F, F)\\} \\cup \\{(F, T)\\} \\cup \\{(T, F)\\} \\cup \\{(T, T)\\}) = 1 \\\\\n",
|
||||
"P((F,F)) + P(\\{(F, T)\\} \\cup \\{(T, F)\\} \\cup \\{(T, T)\\}) = 1 \\\\\n",
|
||||
"0.3 + P(\\{(F, T)\\} \\cup \\{(T, F)\\} \\cup \\{(T, T)\\}) = 1 \\\\\n",
|
||||
"P(\\{(F, T)\\} \\cup \\{(T, F)\\} \\cup \\{(T, T)\\}) = 0.7 \\\\\n",
|
||||
"P(\\{(F, T)\\} \\cup \\{(T, T)\\}) + P((T, F)) = 0.7 \\\\\n",
|
||||
"0.5 + P((T, F)) = 0.7 \\\\\n",
|
||||
"P((T, F)) = 0.2 \\\\\n",
|
||||
"P(\\{(T, F)\\} \\cup \\{(T, T)\\}) + P((F, T)) = 0.7 \\\\\n",
|
||||
"P((F, T)) = 0.1\n",
|
||||
"\\end{align*}\n",
|
||||
"\n",
|
||||
"Find the following probabilities:\n",
|
||||
"\n",
|
||||
"a. The probability that it will rain today or tomorrow.\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"P((T, F) \\cup (F, T) \\cup (T, T)) = 0.7\n",
|
||||
"\\end{align*}\n",
|
||||
"\n",
|
||||
"b. The probability that it will rain today and tomorrow.\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"P((T, T)) = 1 - 0.3 - 0.2 - 0.1 = 0.4\n",
|
||||
"\\end{align*}\n",
|
||||
"\n",
|
||||
"c. The probability that it will rain today but not tomorrow.\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"P((T, F)) = 0.2\n",
|
||||
"\\end{align*}\n",
|
||||
"\n",
|
||||
"d. The probability that it either will rain today or tomorrow, but not both.\n",
|
||||
"\n",
|
||||
"\\begin{align*} \n",
|
||||
"P(\\{(T, F)\\} \\cup \\{(F, T)\\}) = P((T, F)) + P((F, T)) = 0.2 + 0.1 = 0.3\n",
|
||||
"\\end{align*}\n",
|
||||
"\n"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "8b5131dd-5ebd-4156-b808-f8df273317fb",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"Example 1.12 - soln\n",
|
||||
"\n",
|
||||
"$S = \\{ -1, 0, 1, 2, 3, ... \\}$\n",
|
||||
"\n",
|
||||
"$\\forall x \\in S, P(x) = \\frac{1}{2^{x + 2}}$\n",
|
||||
"\n",
|
||||
"What is the probability that I win more than or equal to 1 dollar and less than 4 dollars?\n",
|
||||
"\n",
|
||||
"\\begin{align*} \n",
|
||||
"P({1, 2, 3}) = P(1) + P(2) + P(3) \\\\\n",
|
||||
"= 1/8 + 1/16 + 1/32\n",
|
||||
"\\end{align*}\n",
|
||||
"\n",
|
||||
"What is the probability that I win more than 2 dollars?\n",
|
||||
"\n",
|
||||
"\\begin{align*} \n",
|
||||
"\\sum_{i=3}^{\\infty} P(i) = P(3) + P(4) + P(5) + P(6) + ... \\\\\n",
|
||||
"= 1/32 + 1/64 + 1/128 + 1/256 + ... \\\\\n",
|
||||
"=\\frac{\\frac{1}{32}}{1 - \\frac{1}{2}}\n",
|
||||
"=\\frac{1}{16}\n",
|
||||
"\\end{align*}"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": null,
|
||||
"id": "69962960-18a6-436b-b9eb-9a6dfae7559a",
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": []
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": null,
|
||||
"id": "11163495-a6cc-43b9-bf41-02748b13d210",
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": []
|
||||
}
|
||||
],
|
||||
"metadata": {
|
||||
"kernelspec": {
|
||||
"display_name": "roadmap (3.14.5)",
|
||||
"language": "python",
|
||||
"name": "python3"
|
||||
},
|
||||
"language_info": {
|
||||
"codemirror_mode": {
|
||||
"name": "ipython",
|
||||
"version": 3
|
||||
},
|
||||
"file_extension": ".py",
|
||||
"mimetype": "text/x-python",
|
||||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.14.5"
|
||||
}
|
||||
},
|
||||
"nbformat": 4,
|
||||
"nbformat_minor": 5
|
||||
}
|
||||
BIN
Binary file not shown.
+331083
File diff suppressed because one or more lines are too long
+16
@@ -0,0 +1,16 @@
|
||||
# Introduction to Probability, Statistics, and Random Processes - Hossein Pishro-Nik
|
||||
|
||||
[pdf](./Introduction%20to%20Probability,%20Statistics,%20and%20Random%20Processes%20-%20Hossein%20Pishro-Nik.pdf)
|
||||
|
||||
total pages=1007
|
||||
|
||||
**Currently reading:** chapter 3, page 236
|
||||
|
||||
TODO:
|
||||
|
||||
- 3.2.5 problems
|
||||
- ch3 end of chapter problems
|
||||
- 4.1.4 problems
|
||||
- 4.2.6 problems
|
||||
- 4.3.3 problems
|
||||
- ch4 end of chapter problems
|
||||
+791
File diff suppressed because one or more lines are too long
+222
@@ -0,0 +1,222 @@
|
||||
State De Morgan's Law for $n$ sets
|
||||
---
|
||||
|
||||
For any sets $A_1, A_2, \dots, A_n$:
|
||||
|
||||
$$\left(\bigcup_{i=1}^n A_i\right)^c = \bigcap_{i=1}^n A_i^c$$
|
||||
|
||||
$$\left(\bigcap_{i=1}^n A_i\right)^c = \bigcup_{i=1}^n A_i^c$$
|
||||
|
||||
The complement of a union is the intersection of complements, and vice versa.
|
||||
|
||||
===
|
||||
|
||||
State the Distributive Law for sets $A$, $B$, and $C$
|
||||
---
|
||||
|
||||
$$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$
|
||||
|
||||
$$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$
|
||||
|
||||
===
|
||||
|
||||
State the inclusion-exclusion principle for a finite collection of sets $A_1, A_2, A_3, \dots A_n$ where $n=2$
|
||||
|
||||
---
|
||||
|
||||
$n = 2$ case:
|
||||
|
||||
$|A \cup B| = |A| + |B| - |A \cap B|$
|
||||
|
||||
---
|
||||
|
||||
$n = 3$ case:
|
||||
|
||||
$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$
|
||||
|
||||
---
|
||||
|
||||
For a finite collection of sets $A_1, A_2, A_3, \dots A_n$, we have
|
||||
|
||||
$\left| \bigcup_{i=1}^n A_i \right| = \sum_{i=1}^n |A_i| - \sum_{i < j} |A_i \cap A_j| + \sum_{i < j < k} |A_i \cap A_j \cap A_k| - \dots + (-1)^{n-1} |A_1 \cap A_2 \cap \dots \cap A_n|$
|
||||
|
||||
===
|
||||
|
||||
For the function
|
||||
|
||||
$$f: A \to B$$
|
||||
|
||||
State the following of $f$:
|
||||
|
||||
1. Domain
|
||||
2. Co-domain
|
||||
3. Range
|
||||
|
||||
---
|
||||
|
||||
1. $A$
|
||||
2. $B$
|
||||
3. Set of all possible outputs of $f$ (not necessarily $B$)
|
||||
|
||||
===
|
||||
|
||||
State definition for each the following:
|
||||
|
||||
1. Random experiment
|
||||
2. Outcome
|
||||
3. Sample space
|
||||
4. Event
|
||||
|
||||
---
|
||||
|
||||
1. A **random experiment** is a process by which we observe something uncertain
|
||||
2. An **outcome** is a result of a random experiment
|
||||
3. The **sample space** $S$ is the set of all possible outcomes
|
||||
4. An **event** is a subset of the sample space
|
||||
|
||||
===
|
||||
|
||||
State the Axioms of Probability
|
||||
|
||||
---
|
||||
**Axioms of Probability**
|
||||
|
||||
1. For any event $A$, $P(A) \geq 0$
|
||||
2. $P(S) = 1$
|
||||
3. If $A_1, A_2, A_3, \dots$ are disjoint events, then $P(A_1 \cup A_2 \cup A_3 \cup \dots) = P(A_1) + P(A_2) + P(A_3) + \dots$
|
||||
|
||||
===
|
||||
|
||||
In a finite sample space $S$, where all outcomes are equally likely, what is the probability of any event $A$?
|
||||
|
||||
---
|
||||
|
||||
$P(A) = \frac{|A|}{|S|}$
|
||||
|
||||
===
|
||||
|
||||
What is $P(A^c)$ in terms of $P(A)$?
|
||||
---
|
||||
|
||||
$$P(A^c) = 1 - P(A)$$
|
||||
|
||||
Follows from the axioms: $A$ and $A^c$ are disjoint, and $A \cup A^c = S$, so $P(A) + P(A^c) = P(S) = 1$.
|
||||
|
||||
===
|
||||
|
||||
Define conditional probability $P(A|B)$
|
||||
---
|
||||
|
||||
The probability of $A$ given $B$ (when $P(B) > 0$):
|
||||
|
||||
$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
|
||||
|
||||
In the equally-likely case:
|
||||
|
||||
$$P(A|B) = \frac{|A \cap B|}{|B|}$$
|
||||
|
||||
===
|
||||
|
||||
State the chain rule of probability for $n$ events
|
||||
---
|
||||
|
||||
$$P(A_1 \cap A_2 \cap \cdots \cap A_n) = P(A_1)\,P(A_2|A_1)\,P(A_3|A_1,A_2)\cdots P(A_n|A_1,A_2,\dots,A_{n-1})$$
|
||||
|
||||
Each factor conditions on all previously listed events.
|
||||
|
||||
===
|
||||
|
||||
What conditions must hold for three events $A$, $B$, $C$ to be independent?
|
||||
---
|
||||
|
||||
All four conditions must hold:
|
||||
|
||||
1. $P(A \cap B) = P(A)P(B)$
|
||||
2. $P(A \cap C) = P(A)P(C)$
|
||||
3. $P(B \cap C) = P(B)P(C)$
|
||||
4. $P(A \cap B \cap C) = P(A)P(B)P(C)$
|
||||
|
||||
Pairwise independence alone is **not** sufficient.
|
||||
|
||||
===
|
||||
|
||||
What does it mean for $n$ events $A_1, A_2, \dots, A_n$ to be independent?
|
||||
---
|
||||
|
||||
Every subset of the events must satisfy the product rule. That is, for all $i < j < k < \dots$:
|
||||
|
||||
$$P(A_i \cap A_j) = P(A_i)P(A_j)$$
|
||||
$$P(A_i \cap A_j \cap A_k) = P(A_i)P(A_j)P(A_k)$$
|
||||
$$\vdots$$
|
||||
$$P(A_1 \cap A_2 \cap \cdots \cap A_n) = \prod_{i=1}^n P(A_i)$$
|
||||
|
||||
===
|
||||
|
||||
If $A_1, A_2, \dots, A_n$ are independent, what is $P(A_1 \cup A_2 \cup \cdots \cup A_n)$?
|
||||
---
|
||||
|
||||
$$P(A_1 \cup A_2 \cup \cdots \cup A_n) = 1 - \prod_{i=1}^n (1 - P(A_i))$$
|
||||
|
||||
Derived by taking the complement (none of the events occur) and using independence.
|
||||
|
||||
===
|
||||
|
||||
What is the difference between disjoint and independent?
|
||||
|
||||
---
|
||||
|
||||
- Disjoint: $A$ and $B$ cannot occur at the same time. $A \cap B = \empty$
|
||||
- Independent: $A$ does not give any information about $B$
|
||||
|
||||
===
|
||||
|
||||
State the Law of Total Probability using event $B$ and its complement
|
||||
---
|
||||
|
||||
$$P(A) = P(A|B)\,P(B) + P(A|B^c)\,P(B^c)$$
|
||||
|
||||
===
|
||||
|
||||
State the general Law of Total Probability for a partition of the sample space
|
||||
---
|
||||
|
||||
If $B_1, B_2, B_3, \dots$ is a partition of $S$, then for any event $A$:
|
||||
|
||||
$$P(A) = \sum_i P(A \cap B_i) = \sum_i P(A|B_i)\,P(B_i)$$
|
||||
|
||||
===
|
||||
|
||||
State Bayes' Rule for two events $A$ and $B$
|
||||
---
|
||||
|
||||
For $P(A) \neq 0$:
|
||||
|
||||
$$P(B|A) = \frac{P(A|B)\,P(B)}{P(A)}$$
|
||||
|
||||
===
|
||||
|
||||
State Bayes' Rule when $B_1, B_2, \dots$ form a partition of $S$
|
||||
---
|
||||
|
||||
For any event $A$ with $P(A) \neq 0$:
|
||||
|
||||
$$P(B_j|A) = \frac{P(A|B_j)\,P(B_j)}{\displaystyle\sum_i P(A|B_i)\,P(B_i)}$$
|
||||
|
||||
The denominator expands $P(A)$ via the Law of Total Probability.
|
||||
|
||||
===
|
||||
|
||||
State the meaning of conditionally independent between events $A$ and $B$ given event $C$. Note $P(C) \gt 0$.
|
||||
|
||||
---
|
||||
|
||||
$$P(A \cap B \mid C) = P(A \mid C)P(B \mid C)$$
|
||||
|
||||
===
|
||||
|
||||
State the multiplication rule for $P(A \cap B)$
|
||||
---
|
||||
|
||||
$$P(A \cap B) = P(A|B)\,P(B) = P(B|A)\,P(A)$$
|
||||
|
||||
Rearrangement of the definition of conditional probability.
|
||||
+17
@@ -0,0 +1,17 @@
|
||||
For any event $A$, prove $P(A^c) = 1 - P(A)$
|
||||
|
||||
---
|
||||
|
||||
$$
|
||||
1 = P(S)
|
||||
= P(A \cup A^c)
|
||||
= P(A) + P(A^c)
|
||||
$$
|
||||
|
||||
===
|
||||
|
||||
Prove $P(A \setminus B) = P(A) - P(A \cap B)$
|
||||
|
||||
---
|
||||
|
||||
TODO
|
||||
+216
@@ -0,0 +1,216 @@
|
||||
{
|
||||
"cells": [
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 1,
|
||||
"id": "206bf674",
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": [
|
||||
"import os\n",
|
||||
"import sys\n",
|
||||
"\n",
|
||||
"import matplotlib.pyplot as plt\n",
|
||||
"import numpy as np\n",
|
||||
"import pandas as pd\n",
|
||||
"import seaborn as sns\n",
|
||||
"\n",
|
||||
"sns.set_theme(style=\"whitegrid\", context=\"notebook\")"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "612bd02c",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"# Chapter 1 Summary Notes"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "be70f5df",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Intro (1.0.0 - 1.3.1)\n",
|
||||
"\n",
|
||||
"**Theorem 1.1: De Morgan's law** \\\n",
|
||||
"For any sets $A_1, A_2, A_3, \\dots A_n$, we have\n",
|
||||
"- $(A_1 \\cup A_2 \\cup A_3 \\cup \\dots A_n)^c = A_1^c \\cap A_2^c \\cap A_3^c \\cap \\dots A_n^c$\n",
|
||||
"- $(A_1 \\cap A_2 \\cap A_3 \\cap \\dots A_n)^c = A_1^c \\cup A_2^c \\cup A_3^c \\cup \\dots A_n^c$\n",
|
||||
"\n",
|
||||
"**Theorem 1.2: Distributive law** \\\n",
|
||||
"For any sets $A, B,$ and $C$ we have\n",
|
||||
"- $A \\cap (B \\cup C) = (A \\cap B)\\cup(A \\cap C)$\n",
|
||||
"- $A \\cup (B \\cap C) = (A \\cup B)\\cap(A \\cup C)$\n",
|
||||
"\n",
|
||||
"**Inclusion-exclusion principle** \\\n",
|
||||
"For a finite collection of sets $A_1, A_2, A_3, \\dots A_n$, we have\n",
|
||||
"\n",
|
||||
"$\\left| \\bigcup_{i=1}^n A_i \\right| = \\sum_{i=1}^n |A_i| - \\sum_{i < j} |A_i \\cap A_j| + \\sum_{i < j < k} |A_i \\cap A_j \\cap A_k| - \\dots + (-1)^{n-1} |A_1 \\cap A_2 \\cap \\dots \\cap A_n|$\n",
|
||||
"\n",
|
||||
"$n = 2$ case:\n",
|
||||
"\n",
|
||||
"$|A \\cup B| = |A| + |B| - |A \\cap B|$\n",
|
||||
"\n",
|
||||
"$n = 3$ case:\n",
|
||||
"\n",
|
||||
"$|A \\cup B \\cup C| = |A| + |B| + |C| - |A \\cap B| - |A \\cap C| - |B \\cap C| + |A \\cap B \\cap C|$\n",
|
||||
"\n"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "c7475d4f",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Random experiments (1.3.1 - 1.4)\n",
|
||||
"\n",
|
||||
"- A **random experiment** is a process by which we observe something uncertain\n",
|
||||
"- An **outcome** is a result of a random experiment\n",
|
||||
"- The **sample space** $S$ is the set of all possible outcomes\n",
|
||||
"- An **event** $A$ is any subset of $S$\n",
|
||||
"\n",
|
||||
"> In the context of a random experiment, the sample space is our *universal set*\n",
|
||||
"\n",
|
||||
"**Axioms of Probability**\n",
|
||||
"\n",
|
||||
"1. For any event $A$, $P(A) \\geq 0$\n",
|
||||
"2. $P(S) = 1$\n",
|
||||
"3. If $A_1, A_2, A_3, \\dots$ are disjoint events, then $P(A_1 \\cup A_2 \\cup A_3 \\cup \\dots) = P(A_1) + P(A_2) + P(A_3) + \\dots$\n",
|
||||
"\n",
|
||||
"**Some notation**\n",
|
||||
"\n",
|
||||
"- $P(A \\cap B) = P(A$ and $B) = P(A,B)$\n",
|
||||
"- $P(A \\cup B) = P(A$ or $B)$\n",
|
||||
"\n",
|
||||
"In a finite sample space $S$, where all outcomes are equally likely, the probability of any event $A$ can be found by\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"P(A) = \\frac{|A|}{|S|}\n",
|
||||
"\\end{align*}"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "b705ef32",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Conditional probability (1.4.0)\n",
|
||||
"\n",
|
||||
"If $A$ and $B$ are twos events in sample space $S$, then the **conditional probability of $A$ given $B$** is defined as\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"P(A|B) = \\frac{P(A \\cap B)}{P(B)}, \\text{when } P(B) > 0\n",
|
||||
"\\end{align*}\n",
|
||||
"\n",
|
||||
"or\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"P(A|B) = \\frac{|A \\cap B|}{|B|}, \\text{when } P(B) > 0\n",
|
||||
"\\end{align*}\n",
|
||||
"\n",
|
||||
"For events $A, B,$ and $C$, with $P(C) \\gt 0$, we have\n",
|
||||
"\n",
|
||||
"- $P(A^c|C) = 1 - P(A|C)$\n",
|
||||
"- $P(\\empty|C) = 0$\n",
|
||||
"- $P(A|C) \\leq 1$\n",
|
||||
"- $P(A \\setminus B|C) = P(A|C) - P(A \\cap B|C)$\n",
|
||||
"- $P(A \\cup B|C) = P(A|C) + P(B|C) - P(A \\cap B|C)$\n",
|
||||
"- if $A \\subset B$ then $P(A|C) \\leq P(B|C)$\n",
|
||||
"\n",
|
||||
""
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "188a8fc2",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Independence (1.4.1)\n",
|
||||
"\n",
|
||||
"**Definition.** Two events $A$ and $B$ are *independent* if $P(A \\cap B) = P(A)P(B)$. AKA $P(A|B) = P(A)$\n",
|
||||
"\n",
|
||||
"**Definition.** Three events $A, B,$ and $C$ are independent if **all** of the following conditions hold:\n",
|
||||
"- $P(A \\cap B) = P(A)P(B)$\n",
|
||||
"- $P(A \\cap C) = P(A)P(C)$\n",
|
||||
"- $P(B \\cap C) = P(B)P(C)$\n",
|
||||
"- $P(A \\cap B \\cap C) = P(A)P(B)P(C)$\n",
|
||||
"\n",
|
||||
"**Definition.** $N$ events $A_1, A_2, A_3, \\dots, A_n$ are independent if **all** the following conditions holds:\n",
|
||||
"- $P(A_i \\cap B_j) = P(A_i)P(A_j)$\n",
|
||||
"- $P(A_i \\cap A_j \\cap A_k) = P(A_i)P(A_j)P(A_k)$ where $i \\in [1:n+1]$, $j \\in [i:n+1]$, $k \\in [j:n+1]$\n",
|
||||
"- $\\dots$\n",
|
||||
"- $P(A_1 \\cap A_2 \\cap A_3 \\cap \\dots \\cap A_n) = \\prod_{i=1}^nP(A_i)$\n",
|
||||
"\n",
|
||||
"**Lemma.** \\\n",
|
||||
"If $A$ and $B$ are independent then\n",
|
||||
"- $A$ and $B^c$ are independent\n",
|
||||
"- $A^c$ and $B$ are independent\n",
|
||||
"- $A^c$ and $B^c$ are independent\n",
|
||||
"\n",
|
||||
"**Definition.** If $A_1, A_2, \\dots, A_n$ are independent then\n",
|
||||
"$$P(A_1 \\cup A_2 \\cup \\dots \\cup A_n) = 1 - (1 - P(A_1))(1 - P(A_2))\\dots(1 - P(A_n))$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "90cf5b00",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Law of Total Probability (1.4.2)\n",
|
||||
"\n",
|
||||
"$$P(A) = P(A|B)P(B) + P(A|B^c)P(B^c)$$\n",
|
||||
"\n",
|
||||
"**Definition.** Law of Total Probability: \\\n",
|
||||
"If $B_1, B_2, B_3, \\dots $ is a partition of the sample space $S$, then for any event $A$ we have\n",
|
||||
"\n",
|
||||
"$$P(A) = \\sum_i P(A \\cap B_i) = \\sum_i P(A|B_i)P(B_i)$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "c33df800",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Bayes' Rule (1.4.3)\n",
|
||||
"\n",
|
||||
"**Definition.** Bayes' Rule\n",
|
||||
"\n",
|
||||
"- For any two events $A$ and $B$, where $P(A) \\neq 0$, we have\n",
|
||||
"\n",
|
||||
"$$P(B|A) = \\frac{P(A|B)P(B)}{P(A)}$$\n",
|
||||
"\n",
|
||||
"- If $B_1, B_2, B_3, \\dots$ form a partition of the sample space $S$, and $A$ is any event with $P(A) \\neq 0$, we have\n",
|
||||
"\n",
|
||||
"$$P(B_j|A) = \\frac{P(A|B_j)P(B_j)}{\\sum_i P(A|B_i)P(B_i)}$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "849d7c99",
|
||||
"metadata": {},
|
||||
"source": []
|
||||
}
|
||||
],
|
||||
"metadata": {
|
||||
"kernelspec": {
|
||||
"display_name": "roadmap (3.14.5)",
|
||||
"language": "python",
|
||||
"name": "python3"
|
||||
},
|
||||
"language_info": {
|
||||
"codemirror_mode": {
|
||||
"name": "ipython",
|
||||
"version": 3
|
||||
},
|
||||
"file_extension": ".py",
|
||||
"mimetype": "text/x-python",
|
||||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.14.5"
|
||||
}
|
||||
},
|
||||
"nbformat": 4,
|
||||
"nbformat_minor": 5
|
||||
}
|
||||
+139
@@ -0,0 +1,139 @@
|
||||
{
|
||||
"cells": [
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 2,
|
||||
"id": "c58309b2",
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": [
|
||||
"import os\n",
|
||||
"import sys\n",
|
||||
"\n",
|
||||
"import matplotlib.pyplot as plt\n",
|
||||
"import numpy as np\n",
|
||||
"import pandas as pd\n",
|
||||
"import seaborn as sns\n",
|
||||
"\n",
|
||||
"sns.set_theme(style=\"whitegrid\", context=\"notebook\")"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "a6732353-51d5-4478-9cf8-5834e57e5a4e",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"# Chapter 2 Notes"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "9f0046c2",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Counting (2.0.0 - 2.1.5)\n",
|
||||
"\n",
|
||||
"***Definition.*** Multiplication Principle: \\\n",
|
||||
"Suppose that we perform $r$ experiments such that the $k\\text{th}$ experiment has $n_k$ possible outcomes, for $k=1,2,\\dots,r$. Then there are a total of $n_1 \\times n_2 \\times n_3 \\times \\dots \\times n_r$ possible outcomes for the sequence of $r$ experiments\n",
|
||||
"\n",
|
||||
"### Terminology\n",
|
||||
"\n",
|
||||
"- **Sampling**: Sampling from a set means choosing an element from that set. We\n",
|
||||
"often **draw** a sample at random from a given set in which each element of the\n",
|
||||
"set has equal chance of being chosen\n",
|
||||
"\n",
|
||||
"- **With or without replacement**: Usually we draw multiple samples from a set. If\n",
|
||||
"we put each object back after each draw, we call this sampling with\n",
|
||||
"replacement. In this case a single object can be possibly chosen multiple times.\n",
|
||||
"For example, if A = {a1, a2, a3, a4} and we pick 3 elements with replacement, a\n",
|
||||
"possible choice might be (a3, a1, a3). Thus \"with replacement\" means \"repetition\n",
|
||||
"is allowed.\" On the other hand, if repetition is not allowed, we call it sampling\n",
|
||||
"without replacement\n",
|
||||
"\n",
|
||||
"- **Ordered or unordered**: If ordering matters (i.e.: a1, a2, a3 ≠ a2, a3, a1), this is\n",
|
||||
"called ordered sampling. Otherwise, it is called unordered\n",
|
||||
"\n",
|
||||
"### Counting Formulas\n",
|
||||
"\n",
|
||||
"- **ordered sampling with replacement:** $n^k$\n",
|
||||
"\n",
|
||||
"- **ordered sampling without replacement:** $n$ permute $k$ $\\quad$ ie $P^n_k = \\frac{n!}{(n - k)!}$\n",
|
||||
"\n",
|
||||
"- **unordered sampling without replacement:** $n$ choose $k$ $\\quad$ ie $\\binom{n}{k} = \\frac{n!}{k!(n-k)!}$\n",
|
||||
"\n",
|
||||
"- **unordered sampling with replacement:** $\\binom{n + k - 1}{k}$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "2e93e0fe",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Problem Solving Principles"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "87279e59",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"When solving a combinatorics problem, consider:\n",
|
||||
"1. Does order matter?\n",
|
||||
" - Yes → Permutations\n",
|
||||
" - No → Combinations\n",
|
||||
"\n",
|
||||
" - \"Are HHHTT and THHHT the same outcome to me?\"\n",
|
||||
"\n",
|
||||
"2. Are we sampling with of without replacement?\n",
|
||||
" - Without replacement → Hypergeometric (phone problem)\n",
|
||||
" - With replacement → Binomial (coin flips)\n",
|
||||
" \n",
|
||||
" \"Can the same item be chosen twice?\"\n",
|
||||
"3. Are the \"groups\" labeled or unlabeled?\n",
|
||||
" - Labeled/distinguishable → Just multiply combinations\n",
|
||||
" - Unlabeled/interchangeable → Divide by k!\n",
|
||||
"\n",
|
||||
" \"Does it matter which group is called group 1?\"\n",
|
||||
"4. Are the items distinguishable?\n",
|
||||
" - Distinguishable → Each item is unique, classical probability applies\n",
|
||||
" - Indistinguishable → Outcomes are not equally likely, be careful\n",
|
||||
"\n",
|
||||
" \"Could I label these items 1 to n?\"\n",
|
||||
"5. Is complement of inclusion-exclusion easier?\n",
|
||||
" - Complement → When \"at least\" or \"at most\" language appears\n",
|
||||
" - Inclusion-Exclusion → When events overlap\n",
|
||||
"\n",
|
||||
" \"Is the opposite event simpler to count?\"\n",
|
||||
"6. Am I counting each outcome exactly once? \n",
|
||||
" - If yes, done. Otherwhise we are overcounting or undercounting"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "9c5783dc",
|
||||
"metadata": {},
|
||||
"source": []
|
||||
}
|
||||
],
|
||||
"metadata": {
|
||||
"kernelspec": {
|
||||
"display_name": "roadmap (3.14.5)",
|
||||
"language": "python",
|
||||
"name": "python3"
|
||||
},
|
||||
"language_info": {
|
||||
"codemirror_mode": {
|
||||
"name": "ipython",
|
||||
"version": 3
|
||||
},
|
||||
"file_extension": ".py",
|
||||
"mimetype": "text/x-python",
|
||||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.14.5"
|
||||
}
|
||||
},
|
||||
"nbformat": 4,
|
||||
"nbformat_minor": 5
|
||||
}
|
||||
+328
File diff suppressed because one or more lines are too long
+601
File diff suppressed because one or more lines are too long
+101
@@ -0,0 +1,101 @@
|
||||
{
|
||||
"cells": [
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 1,
|
||||
"id": "206bf674",
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": [
|
||||
"import os\n",
|
||||
"import sys\n",
|
||||
"\n",
|
||||
"import matplotlib.pyplot as plt\n",
|
||||
"import numpy as np\n",
|
||||
"import pandas as pd\n",
|
||||
"import seaborn as sns\n",
|
||||
"\n",
|
||||
"sns.set_theme(style=\"whitegrid\", context=\"notebook\")"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "612bd02c",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"# Chapter 4 Example Problems"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "5ce47b88",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"4.2 \n",
|
||||
"\n",
|
||||
"a. \n",
|
||||
"\n",
|
||||
"$\\int_{-\\infty}^{\\infty}f_X(u)du = 1$\n",
|
||||
"\n",
|
||||
"ie\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"\\int_{-\\infty}^{0^+}f_X(u)du + \\int_{0}^{\\infty}f_X(x)dx = 1 \\\\\n",
|
||||
"0 + \\int_{0}^{\\infty}ce^{-x}dx = 1 \\\\\n",
|
||||
"c\\cdot\\lim_{t \\to \\infty }\\int_{0}^{t}e^{-x}dx = 1 \\\\\n",
|
||||
"c\\cdot\\lim_{t \\to \\infty } [-e^{-x}]_0^t = 1 \\\\\n",
|
||||
"c\\cdot\\lim_{t \\to \\infty } ((-e^{-t}) - (-e^{-0})) = 1 \\\\\n",
|
||||
"c\\cdot\\lim_{t \\to \\infty } (-e^{-t} + 1) = 1 \\\\\n",
|
||||
"c\\cdot 1 = 1 \\\\\n",
|
||||
"c = 1\n",
|
||||
"\\end{align*}\n",
|
||||
"\n",
|
||||
"b.\n",
|
||||
"\n",
|
||||
"Note that\n",
|
||||
"\n",
|
||||
"$$F_X(x) = \\int_{-\\infty}^x f_X(u)du$$\n",
|
||||
"\n",
|
||||
"and $c=1$\n",
|
||||
"\n",
|
||||
"so\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"F_X(x) = \\begin{cases} \n",
|
||||
"1-e^{-x} & \\text{for } x \\geq 0 \\\\ \n",
|
||||
"0 & \\text{otherwise} \n",
|
||||
"\\end{cases}\n",
|
||||
"\\end{align*}\n",
|
||||
"\n",
|
||||
"c. $F_X(3) - F_X(1) = $"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "95a3b21c",
|
||||
"metadata": {},
|
||||
"source": []
|
||||
}
|
||||
],
|
||||
"metadata": {
|
||||
"kernelspec": {
|
||||
"display_name": "roadmap (3.14.5)",
|
||||
"language": "python",
|
||||
"name": "python3"
|
||||
},
|
||||
"language_info": {
|
||||
"codemirror_mode": {
|
||||
"name": "ipython",
|
||||
"version": 3
|
||||
},
|
||||
"file_extension": ".py",
|
||||
"mimetype": "text/x-python",
|
||||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.14.5"
|
||||
}
|
||||
},
|
||||
"nbformat": 4,
|
||||
"nbformat_minor": 5
|
||||
}
|
||||
+176
@@ -0,0 +1,176 @@
|
||||
{
|
||||
"cells": [
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 2,
|
||||
"id": "c58309b2",
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": [
|
||||
"import os\n",
|
||||
"import sys\n",
|
||||
"\n",
|
||||
"import matplotlib.pyplot as plt\n",
|
||||
"import numpy as np\n",
|
||||
"import pandas as pd\n",
|
||||
"import seaborn as sns\n",
|
||||
"\n",
|
||||
"sns.set_theme(style=\"whitegrid\", context=\"notebook\")"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "a6732353-51d5-4478-9cf8-5834e57e5a4e",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"# Chapter 4 Notes"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "9f0046c2",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Continuous Random Variables and their Distributions (4.1.0)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "ea9b1f96",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"***Definition*** A random variable $X$ with CDF $F_X(x)$ is said to be continuous if $F_X(x)$ is a continuous for all $x \\in \\mathbb{R}$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "a96da4d3",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Probability Density Function (PDF) (4.1.1)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "c62129a5",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"***Definition*** Consider a continuous random variable $X$ with an absolutely continuous CDF $F_X(x)$. The function $f_X(x)$ defined by\n",
|
||||
"\n",
|
||||
"$$f_X(x) = \\frac{dF_X(x)}{dx} = F'_X(x) \\quad \\text{if } F_X(x) \\text{ is differentiable at } x$$\n",
|
||||
" \n",
|
||||
"is called the probability density function (PDF) of $X$."
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "df411869",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"NOTE: The PDF being constant implies uniformity\n",
|
||||
"\n",
|
||||
"NOTE: For small values of $\\delta$,\n",
|
||||
"\n",
|
||||
"$$P(x \\lt X \\leq x + \\delta) \\approx f_X(x)\\delta$$\n",
|
||||
"\n",
|
||||
"Thus if $f_X(x_1) \\gt f_X(x_2)$, we can say $P(x_1 \\lt X \\leq x_1 + \\delta) \\gt P(x_2 \\lt X \\leq x_2 + \\delta)$, ie the value of $X$ is more likely to be around $x_1$ then $x_2$\n",
|
||||
"\n",
|
||||
"NOTE: The CDF can be obtained from the PDF via (assuming absolute continuity)\n",
|
||||
"\n",
|
||||
"$$F_X(x) = \\int_{-\\infty}^x f_X(u)du$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "f29a3bfb",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"***Properties*** Consider a continuous random variable $X$ with PDF $f_X(x)$. We have\n",
|
||||
"- $f_X(x) \\geq 0, \\forall x \\in \\mathbb{R}$\n",
|
||||
"- $\\int_{-\\infty}^{\\infty}f_X(u)du = 1$\n",
|
||||
"- $P(a \\lt X \\leq b) = F_X(b) - F_X(a) = \\int_a^bf_X(u)du$\n",
|
||||
"- For a set $A$, $P(X \\in A) = \\int_Af_X(u)du$. However, set $A$ must satisfy:"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "6f546fed",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"***Definition*** If $X$ is a continuous random variable, we can write the range of $X$ as\n",
|
||||
"\n",
|
||||
"$$R_X = \\{ x \\mid f_X(x) \\gt 0 \\}$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "170db3a0",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"***Property*** The expected value if a continuous random variable $X$ is\n",
|
||||
"\n",
|
||||
"$$E[X] = \\int_{-\\infty}^{\\infty}xf_X(x)dx$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "deccea51",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"***Property*** Law of the unconscious statistician (LOTUS) for continuous random variables\n",
|
||||
"\n",
|
||||
"$$E[g(X)] = \\int_{-\\infty}^{\\infty}g(x)f_X(x)dx$$\n"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "c0a20195",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"***Property*** Variance for a continuous random variable $X$, we can write\n",
|
||||
"\n",
|
||||
"\\begin{align*}\n",
|
||||
"\\text{Var}(X) &= E[(X - E[x])^2] = \\int_{-\\infty}^{\\infty}(x - E[X])^2f_X(x)dx \\\\\n",
|
||||
"&= E[X^2] - E[X]^2 = \\int_{-\\infty}^{\\infty}x^2f_X(x)dx - E[X]^2\n",
|
||||
"\\end{align*}"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "6f2f6b72",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Functions of Continuous Random Variables"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "f4a95738",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"***Theorem***"
|
||||
]
|
||||
}
|
||||
],
|
||||
"metadata": {
|
||||
"kernelspec": {
|
||||
"display_name": "roadmap (3.14.5)",
|
||||
"language": "python",
|
||||
"name": "python3"
|
||||
},
|
||||
"language_info": {
|
||||
"codemirror_mode": {
|
||||
"name": "ipython",
|
||||
"version": 3
|
||||
},
|
||||
"file_extension": ".py",
|
||||
"mimetype": "text/x-python",
|
||||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.14.5"
|
||||
}
|
||||
},
|
||||
"nbformat": 4,
|
||||
"nbformat_minor": 5
|
||||
}
|
||||
BIN
Binary file not shown.
|
After Width: | Height: | Size: 35 KiB |
Reference in New Issue
Block a user