03_module.ipynb

{
 "cells": [
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Complex Probability Problems\n",
    "\n",
    "Some complex problems we will be able to solve at the end of this session\n",
    "\n",
    "\n",
    "+ What is the probability that it takes three flips or more for a coin to land heads up?\n",
    "+ What is the probability of a coin landing heads up 18 times in a row?\n",
    "+ What is the probability of getting at least one 6 in four throws of a single six-sided die?\n",
    "+ What is the probability of getting at least one double-six in 24 throws of two six-sided dice?\n",
    "+ What is the probability of getting four aces in a row when drawing cards from a standard 52-card deck?\n"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Task\n",
    "\n",
    "+  An advertisement company runs a quick test and shows two ads on the same web page (ad \"A\" and ad \"B\") to 100 users. At the end of the trial, they found:\n",
    "    + 12 users clicked on ad \"A\"\n",
    "    + 17 users clicked on ad \"B\"\n",
    "    + 3 users clicked on both ad \"A\" and ad \"B\"\n",
    "\n",
    "+ Find:\n",
    "    + The empirical probability that a user clicks on ad \"A.\" Assign your result to p_a.\n",
    "    + The empirical probability that a user clicks on ad \"B.\" Assign your result to p_b.\n",
    "    + The empirical probability that a user clicks on both ad \"A\" and ad \"B.\" Assign your result to p_a_and_b.\n",
    "    + The probability that a user clicks on either ad \"A\" or ad \"B.\" Assign your result to p_a_or_b. For this exercise, keep in mind a user can click on both ads, so the events are not mutually exclusive — use the addition rule.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_a = 12/100"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_b=17/100"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_a_and_b=3/100"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_a_or_b= p_a + p_b - p_a_and_b"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Opposite Events"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A company that develops a time-tracking tool sells two kinds of subscription: basic and premium. When a new user tries the product, there's a 0.2 probability the user buys the basic subscription and 0.15 he buys premium. Find:\n",
    "\n",
    "+ The probability that a new user doesn't buy a basic subscription (you'll need to use the P(E) = 1 - P(non-E) formula). Assign your result to p_non_basic.\n",
    "+ The probability that a new user doesn't buy a premium subscription. Assign your result to p_non_premium.\n",
    "+ The probability that a user buys either basic or premium. Assign your result to p_subscription (assume buying basic and buying premium are mutually exclusive).\n",
    "    The probability that a new user doesn't buy a subscription. Assign your result to p_non_subscription.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_non_basic = 1 - 0.2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_non_premium= 1 - 0.15"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_subscription = 0.2 + 0.15"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_non_subscription = 1 - p_subscription"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Set Compliments\n",
    "\n",
    "An advertisement company monitors the activity for a specific ad and shows it repeatedly to the same users (so a single user sees the ad multiple times). Regardless of the number of times the ad is shown to a user, the probability that the user clicks on the ad is 0.5. Find:\n",
    "\n",
    "+ The probability that a user doesn't click on the ad. Assign your answer to p_non_click.\n",
    "+ The probability that it takes two times or less for a user to click on the ad. Assign your answer to p_two_or_less.\n",
    "+ The probability that it takes three times or more for a user to click on the ad. Assign your answer to p_three_or_more.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_non_click = 1 - 0.5\n",
    "p_two_or_less = 3/4\n",
    "p_three_or_more = 1 - 3/4"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Multiplication Rule in Probability\n",
    "\n",
    "What is the Multiplication Rule of Probability? \n",
    "\n",
    "> According to the multiplication rule of probability, the probability of occurrence of both the events A and B is equal to the product of the probability of B occurring and the conditional probability that event A occurring given that event B occurs."
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Task\n",
    "\n",
    "For rolling a fair six-sided die, find:\n",
    "\n",
    "1. The probability of getting a 6 two times in a row. Assign your result to p_6_6.\n",
    "2. The probability of getting a 3 on the first throw and a 2 on the second throw. Assign your result to p_3_2.\n",
    "3. The probability of getting an even number on both throws. Assign your result to p_even_even.\n",
    "4. The probability of getting a 1 on the first throw and an even number on the second throw. Assign your result to p_1_even.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_6_6 = 1/6 * 1/6"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_3_2 = 1/6 * 1/6"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_even_even = 3/6 * 3/6"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_1_even = 1/6 * 3/6"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Task \n",
    "\n",
    "Find the probability of:\n",
    "\n",
    "+ Getting heads up 18 times in a row when flipping a fair coin. Assign your answer to p_18h.\n",
    "+ Getting a six three times in a row when throwing a fair six-sided die. Assign your answer to p_666.\n",
    "+ Not getting any six when throwing a fair six-sided die four times. Assign your answer to p_not_6.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_18h = (1/2)**18\n",
    "p_666 = (1/6)**3\n",
    "p_not_6 = (5/6)**4"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Find the probability of getting at least one double-six in 24 throws of two six-sided dice (the two dice are thrown simultaneously). Assign your answer to p_one_double_6."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_one_double_6 = 1 - (35/36)**24"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Sampling with(out) replacement\n"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " + Getting two kings in a row. Assign your answer to p_kk.\n",
    " + Getting a seven of hearts, followed by a queen of diamonds. Assign your answer to p_7q.\n",
    " + Getting a jack, followed by a queen of diamonds, followed by a king, followed by another jack. Assign your answer to p_jqkj. This one is a bit tricky, so pay attention to the details of the question.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [],
   "source": [
    "p_kk = 4/52 * 3/51\n",
    "p_7q = 1/52 * 1/51\n",
    "p_jqkj = 4/52 * 1/51 * 4/50 * 3/49"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "base",
   "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": "3.9.12 (main, Apr  5 2022, 06:56:58) \n[GCC 7.5.0]"
  },
  "orig_nbformat": 4,
  "vscode": {
   "interpreter": {
    "hash": "d0028ce20b428eb90286e675cfc9d586ae702ba0463c3d3b035a87d9639116bc"
   }
  }
 },
 "nbformat": 4,
 "nbformat_minor": 2
}