{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 34 : Probability And Distributions"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.1, page no. 830"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "From the principle of counting, the required no. of ways are 12∗11∗10∗9= \n",
      "11880\n"
     ]
    }
   ],
   "source": [
    "print \"From the principle of counting, the required no. of ways are 12∗11∗10∗9= \"\n",
    "print 12*11*10*9"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.2.1, page no. 831"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "no. of permutations=9!/(2!∗2!∗2!)\n",
      "45360\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "print \"no. of permutations=9!/(2!∗2!∗2!)\"\n",
    "print math.factorial(9)/(math.factorial(2)*math.factorial(2)*math.factorial(2))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.2.2, page no. 831"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "no. of permutations=9!/(2!∗2!∗3!*3!)\n",
      "2520\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "print \"no. of permutations=9!/(2!∗2!∗3!*3!)\"\n",
    "print math.factorial(9)/(math.factorial(2)*math.factorial(2)*math.factorial(3)*math.factorial(3))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.3.1, page no. 832"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "no. of committees=C(6,3)∗C(5,2)=’\n",
      "200\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"no. of committees=C(6,3)∗C(5,2)=’\"\n",
    "print C(6,3)*C(5,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.3.2, page no. 832"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "no. of committees=C(4,1)∗C(5,2)=’\n",
      "40\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"no. of committees=C(4,1)∗C(5,2)=’\"\n",
    "print C(4,1)*C(5,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.3.3, page no. 833"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "no. of committees=C(6,3)∗C(4,2)=’\n",
      "120\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"no. of committees=C(6,3)∗C(4,2)=’\"\n",
    "print C(6,3)*C(4,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.4.1, page no. 834"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The probability of getting a four is 1/6=  0.166666666667\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "print \"The probability of getting a four is 1/6= \",1./6"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.4.2, page no. 834"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The probability of getting an even no. 1/2=  0.5\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "print \"The probability of getting an even no. 1/2= \",1./2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.5, page no. 835"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The probability of 53 Sundays is 2/7=  0.285714285714\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "print \"The probability of 53 Sundays is 2/7= \",2./7"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.6, page no. 835"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The five digits can be arranged in 5! ways =  120\n",
      "Of which 4! will begin with 0 =  24\n",
      "So, total no. of five digit numbers = 5!−4! =  96\n",
      "The numbers ending in 04, 12, 20, 24, 32, 40 will be divisible by 4 \n",
      "numbers ending in 04 = 3! =  6\n",
      "numbers ending in 12 = 3!−2! =  4\n",
      "numbers ending in 20 = 3! =  6\n",
      "numbers ending in 24 = 3!−2! =  4\n",
      "numbers ending in 32 = 3!−2! =  4\n",
      "numbers ending in 40 = 3! =  6\n",
      "So, total no. of favourable ways = 6+4+6+4+4+6 =  30\n",
      "probability = 30/96 =  0.3125\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "print \"The five digits can be arranged in 5! ways = \",math.factorial(5)\n",
    "print \"Of which 4! will begin with 0 = \",math.factorial(4)\n",
    "print \"So, total no. of five digit numbers = 5!−4! = \",math.factorial(5)-math.factorial(4)\n",
    "print \"The numbers ending in 04, 12, 20, 24, 32, 40 will be divisible by 4 \"\n",
    "print \"numbers ending in 04 = 3! = \",math.factorial(3)\n",
    "print \"numbers ending in 12 = 3!−2! = \",math.factorial(3)-math.factorial(2)\n",
    "print \"numbers ending in 20 = 3! = \",math.factorial(3)\n",
    "print \"numbers ending in 24 = 3!−2! = \",math.factorial(3)-math.factorial(2)\n",
    "print \"numbers ending in 32 = 3!−2! = \",math.factorial(3)-math.factorial(2)\n",
    "print \"numbers ending in 40 = 3! = \",math.factorial(3)\n",
    "print \"So, total no. of favourable ways = 6+4+6+4+4+6 = \",6+4+6+4+4+6\n",
    "print \"probability = 30/96 = \",30./96"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.7, page no. 836"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total no. of possible cases = C(40,4) =  91390\n",
      "Favourable outcomes = C(24,2)∗C(15,1) =  4140\n",
      "Probability =  0.0453003610898\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"Total no. of possible cases = C(40,4) = \",C(40,4)\n",
    "print \"Favourable outcomes = C(24,2)∗C(15,1) = \",C(24,2)*C(15,1)\n",
    "print \"Probability = \",float(C(24,2)*C(15,1))/C(40,4)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.8, page no. 836"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total no. of possible cases = C(15,8) =  6435\n",
      "Favourable outcomes = C(5,2)∗C(10,6) =  2100\n",
      "Probability =  0.32634032634\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"Total no. of possible cases = C(15,8) = \",C(15,8)\n",
    "print \"Favourable outcomes = C(5,2)∗C(10,6) = \",C(5,2)*C(10,6)\n",
    "print \"Probability = \",float(C(5,2)*C(10,6))/C(15,8)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.9.1, page no. 837"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total no. of possible cases = C(9,3) =  84\n",
      "Favourable outcomes = C(2,1)∗C(3,1)*C(4,1) =  24\n",
      "Probability =  0.285714285714\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"Total no. of possible cases = C(9,3) = \",C(9,3)\n",
    "print \"Favourable outcomes = C(2,1)∗C(3,1)*C(4,1) = \",C(2,1)*C(3,1)*C(4,1)\n",
    "print \"Probability = \",float(C(2,1)*C(3,1)*C(4,1))/C(9,3)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.9.2, page no. 837"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total no. of possible cases = C(9,3) =  84\n",
      "Favourable outcomes = C(2,2)∗C(7,1)+C(3,2)∗C(6,1)+C(4,2)∗C(5,1) =  55\n",
      "Probability =  0.654761904762\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"Total no. of possible cases = C(9,3) = \",C(9,3)\n",
    "print \"Favourable outcomes = C(2,2)∗C(7,1)+C(3,2)∗C(6,1)+C(4,2)∗C(5,1) = \",C(2,2)*C(7,1)+C(3,2)*C(6,1)+C(4,2)*C(5,1)\n",
    "print \"Probability = \",float(C(2,2)*C(7,1)+C(3,2)*C(6,1)+C(4,2)*C(5,1))/C(9,3)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.9.3, page no. 838"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total no. of possible cases = C(9,3) =  84\n",
      "Favourable outcomes = C(3,3)+C(4,3) =  5\n",
      "Probability =  0.0595238095238\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"Total no. of possible cases = C(9,3) = \",C(9,3)\n",
    "print \"Favourable outcomes = C(3,3)+C(4,3) = \",C(3,3)+C(4,3)\n",
    "print \"Probability = \",5./84"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.13, page no. 840"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of drawing an ace or spade or both from pack of 52 cards = 4/52+13/52−1/52=  17\n"
     ]
    }
   ],
   "source": [
    "print \"Probability of drawing an ace or spade or both from pack of 52 cards = 4/52+13/52−1/52= \",4+13-1/52"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.14.1, page no. 841"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of first card being a king = 4/52 =  0.0769230769231\n",
      "Probability of second card being a queen = 4/52 =  0.0769230769231\n",
      "Probability of drawing both cards in succession = 4/52∗4/52=  0.00591715976331\n"
     ]
    }
   ],
   "source": [
    "print \"Probability of first card being a king = 4/52 = \",4./52\n",
    "print \"Probability of second card being a queen = 4/52 = \",4./52\n",
    "print \"Probability of drawing both cards in succession = 4/52∗4/52= \",(4./52)*(4./52)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.15.1, page no. 842"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of getting 7 in first toss and not getting it in second toss = 1/6∗5/6 =  0.138888888889\n",
      "Probability of not getting 7 in first toss and getting it in second toss = 5/6∗1/6 =  0.138888888889\n",
      "Required probability = 1/6∗5/6+5/6∗1/6 =  0.277777777778\n"
     ]
    }
   ],
   "source": [
    "print \"Probability of getting 7 in first toss and not getting it in second toss = 1/6∗5/6 = \",(1./6)*(5./6)\n",
    "print \"Probability of not getting 7 in first toss and getting it in second toss = 5/6∗1/6 = \",(5./6)*(1./6)\n",
    "print \"Required probability = 1/6∗5/6+5/6∗1/6 = \",((1./6)*(5./6))+((5./6)*(1./6))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.15.2, page no. 842"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of not getting 7 in either toss = 5/6∗5/6 =  0.694444444444\n",
      "Probability of getting 7 at least once = 1−5/6∗5/6 =  0.305555555556\n"
     ]
    }
   ],
   "source": [
    "print \"Probability of not getting 7 in either toss = 5/6∗5/6 = \",(5./6)*(5./6)\n",
    "print \"Probability of getting 7 at least once = 1−5/6∗5/6 = \",1-(5./6)*(5./6)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.15.3, page no. 842"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of getting 7 twice = 1/6∗1/6 =  0.0277777777778\n"
     ]
    }
   ],
   "source": [
    "print \"Probability of getting 7 twice = 1/6∗1/6 = \",(1./6)*(1./6)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.16, page no. 843"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of engineering subject being choosen = (1/3∗3/8)+(2/3∗5/8) =  0.541666666667\n"
     ]
    }
   ],
   "source": [
    "print \"Probability of engineering subject being choosen = (1/3∗3/8)+(2/3∗5/8) = \",((1./3)*(3./8)) +((2./3)*(5./8))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.17, page no. 844"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 36,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of white ball being choosen = 2/6∗6/13+4/6∗5/13 =  0.410256410256\n"
     ]
    }
   ],
   "source": [
    "print \"Probability of white ball being choosen = 2/6∗6/13+4/6∗5/13 = \",((2./6)*(6./13))+((4./6)*(5./13))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.18, page no. 844"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 38,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Chances of winning of A=1/2+(1/2)ˆ2∗(1/2)+(1/2)ˆ4∗(1/2)+(1/2)ˆ6∗(1/2)+..=  0.666666666667\n",
      "Chances of winning of B=1−chances of winning of A =  1\n"
     ]
    }
   ],
   "source": [
    "print \"Chances of winning of A=1/2+(1/2)ˆ2∗(1/2)+(1/2)ˆ4∗(1/2)+(1/2)ˆ6∗(1/2)+..= \",(1./2)/(1-(1./2)**2)\n",
    "print \"Chances of winning of B=1−chances of winning of A = \",1-(2/3)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.19.1, page no. 845"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 40,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total no. of possible outcomes = C(10,2) =  45\n",
      "Favourable outcomes = 5*5 =  25\n",
      "P =  0.555555555556\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"Total no. of possible outcomes = C(10,2) = \",C(10,2)\n",
    "print \"Favourable outcomes = 5*5 = \",5*5\n",
    "print \"P = \",25./45"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.19.2, page no. 845"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total no. of possible outcomes = 10*10 =  100\n",
      "Favourable outcomes = 5*5+5*5 =  50\n",
      "P =  0.5\n"
     ]
    }
   ],
   "source": [
    "print \"Total no. of possible outcomes = 10*10 = \",10*10\n",
    "print \"Favourable outcomes = 5*5+5*5 = \",5*5+5*5\n",
    "print \"P = \",50./100"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.20, page no. 846"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 42,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of A/B = AandB/B =  0.25\n",
      "Probability of B/A = AandB/A =  0.333333333333\n",
      "Probability of AandBnot = A−AandB =  0.166666666667\n",
      "Probability of A/Bnot = AandBnot/Bnot =  0.25\n"
     ]
    }
   ],
   "source": [
    "A = 1./4\n",
    "B = 1./3\n",
    "AorB = 1./2\n",
    "AandB = A+B-AorB\n",
    "print \"Probability of A/B = AandB/B = \",AandB/B\n",
    "print \"Probability of B/A = AandB/A = \",AandB/A\n",
    "print \"Probability of AandBnot = A−AandB = \",A-AandB\n",
    "print \"Probability of A/Bnot = AandBnot/Bnot = \",(1./6)/(1-1./3)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.22, page no. 846"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 43,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of A hitting target = 3/5\n",
      "Probability of B hitting target = 2/5\n",
      "Probability of C hitting target = 3/4\n",
      "Probability that two shots hit = 3/5∗2/5∗(1−3/4)+2/5∗3/4∗(1−3/5)+3/4∗3/5∗(1−2/5) = \n",
      "0.45\n"
     ]
    }
   ],
   "source": [
    "print \"Probability of A hitting target = 3/5\"\n",
    "print \"Probability of B hitting target = 2/5\"\n",
    "print \"Probability of C hitting target = 3/4\"\n",
    "print \"Probability that two shots hit = 3/5∗2/5∗(1−3/4)+2/5∗3/4∗(1−3/5)+3/4∗3/5∗(1−2/5) = \"\n",
    "print (3./5)*(2./5)*(1-3./4)+(2./5)*(3./4)*(1-3./5)+(3./4)*(3./5)*(1-2./5)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.23, page no. 847"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 44,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of problem not getting solved = 1/2∗2/3∗3/4 =  0.25\n",
      "Probability of problem getting solved = 1−(1/2∗2/3∗3/4) =  0.75\n"
     ]
    }
   ],
   "source": [
    "print \"Probability of problem not getting solved = 1/2∗2/3∗3/4 = \",(1./2)*(2./3)*(3./4)\n",
    "print \"Probability of problem getting solved = 1−(1/2∗2/3∗3/4) = \",1-((1./2)*(2./3)*(3./4))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.25, page no. 848"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total frequency=integrate(f,x,0,2) =\n",
      "u1 about origin =  1\n",
      "u2 about origin =  16/15\n",
      "Standard deviation = (u2−u1ˆ2)ˆ0.5=  0.258198889747161\n",
      "Mean deviation about the mean = (1/n)∗(integrate(|x−1|∗(xˆ3),x,0,1)+integrate(|x−1|∗((2−x)ˆ3),x,1,2))\n",
      "1/5\n"
     ]
    }
   ],
   "source": [
    "import sympy\n",
    "\n",
    "print \"Total frequency=integrate(f,x,0,2) =\" \n",
    "n = sympy.integrate('x**3',('x',0,1))+sympy.integrate('(2-x)**3',('x',1,2))\n",
    "print \"u1 about origin = \",\n",
    "u1 = (1/n)*(sympy.integrate('(x)*(x**3)',('x',0,1))+sympy.integrate('(x)*((2-x)**3)',('x',1,2)))\n",
    "print u1\n",
    "print \"u2 about origin = \",\n",
    "u2 = (1/n)*(sympy.integrate('(x**2)*(x**3)',('x',0,1))+sympy.integrate('(x**2)*((2-x)**3)',('x',1,2)))\n",
    "print u2\n",
    "print \"Standard deviation = (u2−u1ˆ2)ˆ0.5= \",(u2-u1**2)**0.5\n",
    "print \"Mean deviation about the mean = (1/n)∗(integrate(|x−1|∗(xˆ3),x,0,1)+integrate(|x−1|∗((2−x)ˆ3),x,1,2))\"\n",
    "print (1/n)*(sympy.integrate('(1-x)*(x**3)',('x',0,1))+sympy.integrate('(x-1)*((2-x)**3)',('x',1,2)))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.26, page no. 849"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 47,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability = (0.45∗0.03)/(0.45∗0.03+0.25∗0.05+0.3∗0.04 =  0.355263157895\n"
     ]
    }
   ],
   "source": [
    "print \"Probability = (0.45∗0.03)/(0.45∗0.03+0.25∗0.05+0.3∗0.04 = \",(0.45*0.03)/(0.45*0.03+0.25*0.05+0.3*0.04)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.27, page no. 849"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 49,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability = (1/3∗2/6∗3/5)/(1/3∗2/6∗3/5+1/3∗1/6∗2/5+1/3∗3/6∗1/5) =  1.05555555556\n"
     ]
    }
   ],
   "source": [
    "print \"Probability = (1/3∗2/6∗3/5)/(1/3∗2/6∗3/5+1/3∗1/6∗2/5+1/3∗3/6∗1/5) = \",\n",
    "print ((1./3)*(2./6)*(3./5))/((1./3)*(2./6)*(3./5))+((1./3)*(1./6)*(2./5))+((1./3)*(3./6)*(1./5))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.28, page no. 850"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 50,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of no success = 8/27 \n",
      "Probability of a success = 1/3 \n",
      "Probability of one success = 4/9\n",
      "Probability of two success = 2/9\n",
      "Probability of three success = 2/9\n",
      "mean=sum of i∗pi =  1.0\n",
      "sum of i∗piˆ2 =  1.66666666667\n",
      "variance = (sum of i∗piˆ2)−1=  0.666666666667\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "\n",
    "print \"Probability of no success = 8/27 \"\n",
    "print \"Probability of a success = 1/3 \"\n",
    "print \"Probability of one success = 4/9\"\n",
    "print \"Probability of two success = 2/9\"\n",
    "print \"Probability of three success = 2/9\"\n",
    "A = numpy.array([[0,1,2,3],[8./27,4./9,2./9,1./27]])\n",
    "print \"mean=sum of i∗pi = \",\n",
    "print A[0,0]*A[1,0]+A[0,1]*A[1,1]+A[0,3]*A[1,3]+A[0,2]*A[1,2]\n",
    "print \"sum of i∗piˆ2 = \",\n",
    "print A[0,0]**2*A[1,0]+A[0,1]**2*A[1,1]+A[0,3]**2*A[1,3]+A[0,2]**2*A[1,2]\n",
    "print \"variance = (sum of i∗piˆ2)−1= \",\n",
    "print A[0,0]**2*A[1,0]+A[0,1]**2*A[1,1]+A[0,3]**2*A[1,3]+A[0,2]**2*A[1,2]-1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.29, page no. 851"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 53,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Sumof all pi = 1 \n",
      "Hence ,\n",
      "p(x<4) =  16.0*k\n",
      "p(x>=5) =  24.0*k\n",
      "p(3<x<=6) =  33.0*k\n",
      "p(x<=2) =  9.0*k\n"
     ]
    }
   ],
   "source": [
    "import sympy,numpy\n",
    "\n",
    "k = sympy.Symbol('k')\n",
    "A =numpy.array([[0,1,2,3,4,5,6],[k,3*k,5*k,7*k,9*k,11*k,13*k]])\n",
    "print \"Sumof all pi = 1 \"\n",
    "print \"Hence ,\"\n",
    "k = 1./49\n",
    "print \"p(x<4) = \",\n",
    "a = A[1,0]+A[1,1]+A[1,3]+A[1,2]\n",
    "print a.evalf()\n",
    "print \"p(x>=5) = \",\n",
    "b = A[1,5]+A[1,6]\n",
    "print b.evalf()\n",
    "print \"p(3<x<=6) = \",\n",
    "c = A[1,4]+A[1,5]+A[1,6]\n",
    "print c.evalf()\n",
    "print \"p(x<=2) = \",\n",
    "e = A[1,0]+A[1,1]+A[1,2]\n",
    "print e.evalf()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.30, page no. 853"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 57,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " Sum of all pi = 1\n",
      "Hence,\n",
      "p(x<6) =  2.0*k**2 + 10.0*k\n",
      "p(x>=6) =  9.0*k**2 + k\n",
      "p(3<x<5) =  8.0*k\n"
     ]
    }
   ],
   "source": [
    "import sympy,numpy,math\n",
    "\n",
    "k = sympy.Symbol('k')\n",
    "A =numpy.array([[0,1,2,3,4,5,6,7],[0,k,2*k,2*k,3*k,k**2,2*k**2,7*k**2+ k]])\n",
    "print \"Sum of all pi = 1\"\n",
    "print \"Hence,\"\n",
    "k = 1./10\n",
    "print \"p(x<6) = \",\n",
    "a = A[1,0]+A[1,1]+A[1,3]+A[1,2]+ A[1,3]+A[1,4]+A[1,6]\n",
    "print a.evalf()\n",
    "print \"p(x>=6) = \",\n",
    "b = A[1,6]+A[1,7]\n",
    "print b.evalf()\n",
    "print \"p(3<x<5) = \",\n",
    "c = A[1,1]+A[1,2]+A[1,3]+A[1,4]\n",
    "print c.evalf()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.31, page no. 853"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 62,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Clearly, f>0 for every x in (1,2) and integrate(f,x,0,numpy.inf)=  1.00000000000000\n",
      "Required probability=p(1<=x<=2)= integrate(f,x,1,2) =  0.232544157934830\n",
      "Cumulative probability function f(2)=integrate(f,x,−%inf,2) =  0.864664716763387\n"
     ]
    }
   ],
   "source": [
    "import sympy,numpy,math\n",
    "y = sympy.Symbol('y')\n",
    "f = math.e**(-y)\n",
    "print \"Clearly, f>0 for every x in (1,2) and integrate(f,x,0,numpy.inf)= \",\n",
    "print sympy.integrate(math.e**(-y),('y',0,sympy.oo))\n",
    "print \"Required probability=p(1<=x<=2)= integrate(f,x,1,2) = \",\n",
    "print sympy.integrate(math.e**(-y),('y',1,2))\n",
    "print \"Cumulative probability function f(2)=integrate(f,x,−%inf,2) = \",\n",
    "print sympy.integrate(math.e**(-y),('y',0,2))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.33, page no. 854"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 65,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total probability = integrate(f,x,0,6)=\n",
      "2*k\n",
      "4*k\n",
      "2*k\n"
     ]
    }
   ],
   "source": [
    "import sympy\n",
    "\n",
    "k = sympy.Symbol('k')\n",
    "print \"Total probability = integrate(f,x,0,6)=\"\n",
    "p = sympy.integrate('k*x',('x',0,2))\n",
    "q = sympy.integrate('2*k',('x',2,4))\n",
    "r = sympy.integrate('-k*x+6*k',('x',4,6))\n",
    "print p\n",
    "print q\n",
    "print r"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 34.34, page no. 854"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "First row of A displays the value of x\n",
      "The second row of x displays the probability of corresponding to x\n",
      "E(x) =  5.5\n",
      "E(x)ˆ2 =  46.5\n",
      "E(2∗x+1)^2=E(4∗xˆ2+4∗x+1) =  209.0\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "\n",
    "A = numpy.array([[-3.,6.,9.],[1./6,1./2,1./3]])\n",
    "print \"First row of A displays the value of x\"\n",
    "print \"The second row of x displays the probability of corresponding to x\"\n",
    "print \"E(x) = \",\n",
    "c = A[0,0]*A[1,0]+A[0,1]*A[1,1]+A[0,2]*A[1,2]\n",
    "print c\n",
    "print \"E(x)ˆ2 = \",\n",
    "b = A[0,0]**2*A[1,0]+A[0,1]**2*A[1,1]+A[0,2]**2*A[1,2]\n",
    "print b\n",
    "print \"E(2∗x+1)^2=E(4∗xˆ2+4∗x+1) = \",4*b+4*c+1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.35, page no. 855"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total frequency=integrate(f,x,0,2)= \n",
      "u1 about origin = \n",
      "u2 about origin = \n",
      "standard deviation = (u2−u1ˆ2)ˆ0.5 =  0.258198889747161\n",
      "Mean deviation about the mean = (1/n)∗(integrate|x−1|∗(xˆ3),x,0,1)+integrate(|x−1|∗((2−x)ˆ3),x,1,2))\n",
      "1/5\n"
     ]
    }
   ],
   "source": [
    "import sympy\n",
    "\n",
    "print \"Total frequency=integrate(f,x,0,2)= \"\n",
    "n = sympy.integrate('x**3',('x',0,1))+sympy.integrate('(2-x)**3',('x',1,2))\n",
    "print \"u1 about origin = \"\n",
    "u1 = (1/n)*(sympy.integrate('(x)*(x**3)',('x',0,1))+sympy.integrate('(x)*((2-x)**3)',('x',1,2)))\n",
    "print \"u2 about origin = \"\n",
    "u2 = (1/n)*(sympy.integrate('(x**2)*(x**3)',('x',0,1))+sympy.integrate('(x**2)*((2-x)**3)',('x',1,2)))\n",
    "print \"standard deviation = (u2−u1ˆ2)ˆ0.5 = \",(u2-u1**2)**0.5\n",
    "print \"Mean deviation about the mean = (1/n)∗(integrate|x−1|∗(xˆ3),x,0,1)+integrate(|x−1|∗((2−x)ˆ3),x,1,2))\"\n",
    "print (1/n)*(sympy.integrate('(1-x)*(x**3)',('x',0,1))+sympy.integrate('(x-1)*((2-x)**3)',('x',1,2)))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.38, page no. 857"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " Probability that exactly two will be defective = C(12,2)∗(0.1)ˆ2∗(0.9)ˆ10 =  0.230127770466\n",
      "Probability that at least two will be defective = 1−(C(12,0)∗(0.9)ˆ12+C(12,1)∗(0.1)∗(0.9)ˆ11) =  0.340997748211\n",
      "The probability that none will be defective = C(12,12)∗(0.9)ˆ12 =  0.282429536481\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"Probability that exactly two will be defective = C(12,2)∗(0.1)ˆ2∗(0.9)ˆ10 = \",C(12,2)*(0.1)**2*(0.9)**10\n",
    "print \"Probability that at least two will be defective = 1−(C(12,0)∗(0.9)ˆ12+C(12,1)∗(0.1)∗(0.9)ˆ11) = \",\n",
    "print 1-(C(12,0)*(0.9)**12+C(12,1)*(0.1)*(0.9)**11)\n",
    "print \"The probability that none will be defective = C(12,12)∗(0.9)ˆ12 = \",C(12,12)*(0.9)**12"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.39, page no. 858"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of 8 heads and 4 tail sin 12 trials = p(8) = C(12,8)∗(1/2)ˆ8∗(1/2)ˆ4=  0.120849609375\n",
      "The expected no. of such cases in 256 sets = 256∗p(8) =  30.9375\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"Probability of 8 heads and 4 tail sin 12 trials = p(8) = C(12,8)∗(1/2)ˆ8∗(1/2)ˆ4= \",C(12.,8.)*(1./2)**8*(1./2)**4\n",
    "print \"The expected no. of such cases in 256 sets = 256∗p(8) = \",256*(495./4096)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 34.40, page no. 859"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of a defective part = 2/20 =0.1 \n",
      "Probability of a non defective part = 0.9 \n",
      "Probabaility of atleast three defectives in a sample =  0.323073194811\n",
      "No. of sample shaving three defective parts = 1000∗0.323 =  323.0\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def C(a,b):\n",
    "    x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
    "    return x\n",
    "print \"Probability of a defective part = 2/20 =0.1 \"\n",
    "print \"Probability of a non defective part = 0.9 \"\n",
    "print \"Probabaility of atleast three defectives in a sample = \",\n",
    "print 1-(C(20.,0.)*(0.9)**20+C(20.,1.)*(0.1)*(0.9)**19+C(20.,2.)*(0.1)**2*(0.9)**18)\n",
    "print \"No. of sample shaving three defective parts = 1000∗0.323 = \",1000*0.323"
   ]
  }
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