{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Lecture 1: Probability and Counting\n", "\n", "## Stat 110, Prof. Joe Blitzstein, Harvard University\n", "\n", "\n", "----" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Definitions\n", "\n", "We start with some basic definitions:\n", "\n", "#### Definition: sample space\n", "\n", "> A __sample space__ is the set of all possible outcomes of an experiment.\n", "\n", "\n", "\n", "#### Definition: event\n", "\n", "> An __event__ is a subset of the sample space.\n", "\n", "\n", "#### Definition: naïve definition of probability\n", "> Under the __naïve definition of probability__, the probability of a given event $A$ occurring is expressed as\n", ">\n", "> \\begin\\{align\\}\n", "> P(A) &= \\frac{ \\text{# favorable outcomes}}{\\text{# possible outcomes}}\n", "> \\end\\{align\\}\n", ">\n", "> assuming all outcomes are equally likely in a finite sample space.\n", "\n", "----" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Counting\n", "\n", "With the __multiplication rule__, if we have an experiment with $n_1$ possible outcomes; and we have a 2<sup>nd</sup> experiment with $n_2$ possible outcomes; ..., and for the r<sup>th</sup> experiment there are $n_r$ possible outcomes; then overall there are $n_1 n_2 ... n_r$ possible outcomes (product).\n", "\n", "Let's say you are ordering ice cream. You can either get a cone or a cup, and the ice cream comes in three flavors. The order of choice here does not matter, and the total number of choices is $2 \\times 3 = 3 \\times 2 = 6$. This can be represented with a very simple [probability tree][1].\n", "\n", "[1]: https://en.wikipedia.org/wiki/Tree_diagram_(probability_theory)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The __binomial coefficient__ is defined as \n", "\n", "\\begin{align}\n", " \\binom{n}{k} =\n", " \\begin{cases}\n", " \\frac{n!}{(n-k)!k!} & \\quad \\text{if } 0 \\le k \\le n \\\\\n", " 0 & \\quad \\text{if } k \\gt n\n", " \\end{cases}\n", "\\end{align}\n", "\n", "This expresses the number of ways you could choose a subset of size $k$ from $n$ items, where order doesn't matter.\n", "\n", "----" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Sampling\n", "\n", "Choose $k$ objects out of $n$\n", "\n", "| | ordered | unordered |\n", "|-----------|:---------:|:-----------:|\n", "| __w/ replacement__ | $n^k$ | ??? |\n", "| __w/o replacement__ | $n(n-1)(n-2) \\ldots (n-k+1)$ | $\\binom{n}{k}$ |\n", "\n", "\n", "* __ordered, w/ replacement__: there are $n$ choices for each $k$, so this follows from the multiplication rule.\n", "* __ordered, w/out replacement__: there are $n$ choices for the 1<sup>st</sup> position; $n-1$ for the 2<sup>nd</sup>; $n-2$ for the 3<sup>rd</sup>; and $n-k+1$ for the $k$<sup>th</sup>.\n", "* __unordered, w/ replacement__: ???\n", "* __unordered, w/out replacement__: the binomial coefficient; think of choosing a hand from a deck of cards.\n", "\n", "Out of the 4 ways of choosing $k$ objects out of $n$, the case of __unordered, with replacement__ is perhaps not as clear-cut and easy to grasp as the other three. Move on to Lecture 2.\n", "\n", "-----" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "View [Lecture 1: Probability and Counting | Statistics 110](http://bit.ly/2vSEEeI) on YouTube." ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "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": "3.6.3" } }, "nbformat": 4, "nbformat_minor": 1 }