Bayesian Statistics The Fun Way - Python Solutions # CHAPTER 04.ipynb

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   "source": [
    "import math\n",
    "import numpy as np\n",
    "from scipy.special import binom as binom_c\n",
    "from scipy.special import comb\n",
    "from scipy.stats import beta\n",
    "from scipy.stats import binom "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# CREATING A BINOMIAL PROBABILITY DISTRIBUTION"
   ]
  },
  {
   "cell_type": "markdown",
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   "source": [
    "Calculating probability of flipping exactly 12 heads in 24 coin tosses"
   ]
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   "source": [
    "general equation: Binomial(k;n,p) = **Binomial_Coefficient(n, k) x p^k x (1 - p)^n-k**\n",
    "\n",
    "where:    \n",
    "  \n",
    "k = number of outcomes we care about  \n",
    "n = number of total trials  \n",
    "p = probability of event happening  "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let's look how we can calculate whole equation using binomial coefficient (binom_c) and other variables, and then using scipy function instead"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 89,
   "metadata": {},
   "outputs": [],
   "source": [
    "p = 0.5\n",
    "n = 24\n",
    "k = 12"
   ]
  },
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   "source": [
    "binom_c(n, k) * (p**k) * ((1 - p)**(n-k))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "which can be calculated with this function as well:\n",
    "\n",
    "Probability Mass Function (PMF)"
   ]
  },
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   "source": [
    "binom.pmf(k=k, n=n, p=p)"
   ]
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   "cell_type": "markdown",
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   "source": [
    "If we need to calculate the probability of interval though, we would use sf function, for instance the probability of getting at least one outcome in 100 trails with probability of 0.00720"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 92,
   "metadata": {},
   "outputs": [],
   "source": [
    "p = 0.00720\n",
    "k = 0\n",
    "n = 100"
   ]
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   "source": [
    "binom.sf(k=k, n=n, p=p)"
   ]
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   "source": [
    "## EXERCISES"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 1. What are the parameters of the binomial distribution for the probability of rolling either a 1 or a 20 on a 20-sided die, if we roll the die 12 times?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 94,
   "metadata": {},
   "outputs": [],
   "source": [
    "p = 0.1\n",
    "n = 12\n",
    "k = 1"
   ]
  },
  {
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   "execution_count": 95,
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     "execution_count": 95,
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   "source": [
    "binom.pmf(k=k, n=n, p=p)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 2. There are four aces in a deck of 52 cards. If you pull a card, return the card, then reshuffle and pull a card again, how many ways can you pull just one ace in five pulls?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 96,
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   "outputs": [
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     "data": {
      "text/plain": [
       "5"
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     "execution_count": 96,
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   "source": [
    "comb(5, 1, exact=True)"
   ]
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   "cell_type": "markdown",
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   "source": [
    "### 3. For example in question 2, what is the probability of pulling five aces in 10 pulls (remember the card is shuffled back in the deck when it is pulled)?"
   ]
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   "cell_type": "code",
   "execution_count": 97,
   "metadata": {},
   "outputs": [],
   "source": [
    "p = 4/52\n",
    "n = 10\n",
    "k = 5"
   ]
  },
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   "source": [
    "binom.pmf(k=k, n=n, p=p)"
   ]
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 4. When you’re searching for a new job, it’s always helpful to have more than one offer on the table so you can use it in negotiations. If you have a 1/5 probability of receiving a job offer when you interview, and you interview with seven companies in a month, what is the probability you’ll have at least two competing offers by the end of that month?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 99,
   "metadata": {},
   "outputs": [],
   "source": [
    "p = 0.2\n",
    "n = 7\n",
    "k = 1"
   ]
  },
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     "execution_count": 100,
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   ],
   "source": [
    "binom.sf(k=k, n=n, p=p)"
   ]
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   "cell_type": "markdown",
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   "source": [
    "### 5. You get a bunch of recruiter emails and find out you have 25 interviews lined up in the next month. Unfortunately, you know this will leave you exhausted, and the probability of getting an offer will drop to 1/10 if you’re tired. You really don’t want to go on this many interviews unless you are at least twice as likely to get at least two competing offers.\n",
    "### Are you more likely to get at least two offers if you go for 25 interviews, or stick to just 7?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 101,
   "metadata": {},
   "outputs": [],
   "source": [
    "p2 = 0.1\n",
    "n2 = 25\n",
    "k2 = 1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 102,
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   "outputs": [
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     },
     "execution_count": 102,
     "metadata": {},
     "output_type": "execute_result"
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   ],
   "source": [
    "binom.sf(k=k2, n=n2, p=p2)"
   ]
  },
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   "cell_type": "code",
   "execution_count": 103,
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     "text": [
      "You should not take it, the probability is only 1.7217647512082521 times higher\n"
     ]
    }
   ],
   "source": [
    "if 2 * (binom.sf(k=k, n=n, p=p)) <= binom.sf(k=k2, n=n2, p=p2):\n",
    "    print('You should should take 25 interviews, for sure!')\n",
    "else:\n",
    "    print(f'You should not take it, the probability is only {(binom.sf(k=k2, n=n2, p=p2))/(binom.sf(k=k, n=n, p=p))}' \n",
    "          ' times higher')"
   ]
  }
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