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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "CHAPTER06:PROBABILITY AND RANDOM VARIABLES"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E07 : Pg 6.16"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Page Number: 6.16\n",
      "# Example 6.7\n",
      "# Given\n",
      "# Pdot=2*Pdash and Pdot+Pdash=1\n",
      "# Therfore, on solving using linear equations\n",
      "import math \n",
      "#a=([1, -2],[1, 1]);\n",
      "##c=([0],[1]);\n",
      "#b=1/(a)*c;\n",
      "Pdash=0.3333333;#b(1,1);\n",
      "Pdot=0.6666667;#b(2,1);\n",
      "print'Pdot:',Pdot\n",
      "print'Pdash:',Pdash"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Pdot: 0.6666667\n",
        "Pdash: 0.3333333\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E14 : Pg 6.18"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Page Number: 6.18\n",
      "# Example 6.14\n",
      "# Given\n",
      "p=0.1;\n",
      "q=0.2;\n",
      "Pm0=0.5;\n",
      "Pr1bym0=p;\n",
      "Pr0bym1=q;\n",
      "# (a) Find Pr0 and Pr1\n",
      "Pm1=1-Pm0;\n",
      "Pr0bym0=1-Pr1bym0;\n",
      "Pr1bym1=1-Pr0bym1;\n",
      "\n",
      "# By formula\n",
      "# P(r0)=(P(r0/m0)*P(m0))+(P(r0/m1)*P(m1);\n",
      "# P(r1)=(P(r1/m0)*P(m0))+(P(r1/m1)*P(m1);\n",
      "Pr0=(Pr0bym0*Pm0)+(Pr0bym1*Pm1);\n",
      "Pr1=(Pr1bym0*Pm0)+(Pr1bym1*Pm1);\n",
      "print'P(r0):',Pr0\n",
      "print'P(r1):',Pr1\n",
      "# (b)P(m0/r0)\n",
      "# Using Bayes Rule\n",
      "# P(m0/r0)=(P(m0)*P(r0/m0)/P(r0))\n",
      "Pm0byr0=(Pm0*Pr0bym0)/Pr0;\n",
      "print'P(m0/r0):',Pm0byr0\n",
      "# (c)P(m1/r1)\n",
      "# Using Bayes Rule\n",
      "# P(m1/r1)=(P(m1)*P(r1/m1)/P(r1))\n",
      "Pm1byr1=(Pm1*Pr1bym1)/Pr1;\n",
      "print'P(m1/r1):',Pm1byr1\n",
      "\n",
      "# (d)Probabilty error\n",
      "# As Pe=(P(r1/m0)*P(m0))+(P(r0/m1)*P(m1))\n",
      "Pe=(Pr1bym0*Pm0)+(Pr0bym1*Pm1);\n",
      "print'Probability error:',Pe\n",
      "\n",
      "# (e)Probabilty that is transmitted correctly\n",
      "# As Pc=(P(r0/m0)*P(m0))+(Pr1bym1*Pm1)\n",
      "\n",
      "Pc=(Pr0bym0*Pm0)+(Pr1bym1*Pm1);\n",
      "print'Probabilty that is transmitted correctly:',Pc"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "P(r0): 0.55\n",
        "P(r1): 0.45\n",
        "P(m0/r0): 0.818181818182\n",
        "P(m1/r1): 0.888888888889\n",
        "Probability error: 0.15\n",
        "Probabilty that is transmitted correctly: 0.85\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E16 : Pg 6.21"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Page Number: 6.21\n",
      "# Example 6.16\n",
      "# Given\n",
      "p=0.4;\n",
      "Pp=p;\n",
      "q=0.3;\n",
      "Pn=q;\n",
      "a=1.; # i start value\n",
      "b=6.; # i end value\n",
      "# (a)Probabilty that all pulses are positive\n",
      "s=1.;\n",
      "for i in range(1,6):\n",
      "    s=s*Pp;\n",
      "print'Probabilty that all pulses are positive:',s\n",
      "\n",
      "# (b)Pulses are positive ,positive, positive, zero,zero,negative\n",
      "Pz=1.-(p+q);\n",
      "s1=(Pp**3.)*(Pz**2.)*Pn;\n",
      "print'Probabilty that all pulses are positive ,positive, positive, zero,zero,negative:',s1"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Probabilty that all pulses are positive: 0.01024\n",
        "Probabilty that all pulses are positive ,positive, positive, zero,zero,negative: 0.001728\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E17 : Pg 6.21"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Page Number: 6.21\n",
      "# Example 6.17\n",
      "# Given\n",
      "import math \n",
      "P1=0.6;\n",
      "P0=0.4;\n",
      "n=5.; # Five digit sequence\n",
      "j=2.; # two outcomes 0 and 1\n",
      "\n",
      "# (a)1,1,0,0,0\n",
      "xf=(math.factorial(n))/((math.factorial(j))*(math.factorial(n-j)));\n",
      "s=xf*(P1**j)*(P0**(n-j));\n",
      "print'Probability for 1,1,0,0,0:',s\n",
      "\n",
      "# (b)atleast 3 1's\n",
      "# P(X>=3)=1-P(X<=2)\n",
      "# Here y=1-x\n",
      "x=0;\n",
      "for k in range(0,2):\n",
      "    f=(math.factorial(n))/((math.factorial(k))*(math.factorial(n-k)));\n",
      "    x=x+(f*((P1**k)*(P0**(n-k))));\n",
      "y=1.-x;\n",
      "print'Probability for atleast three 1 s:',y"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Probability for 1,1,0,0,0: 0.2304\n",
        "Probability for atleast three 1 s: 0.91296\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E20 : Pg 6.23"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Page Number: 6.23\n",
      "# Example 6.20\n",
      "# Given\n",
      "import math \n",
      "pe=0.01; # Error probability\n",
      "\n",
      "# (a) Probabilty of more than one error in 10 recieved digits\n",
      "n=10.;\n",
      "# As P(X>1)=1-P(X=0)-P(X=1)\n",
      "# Let x=P(X>1)\n",
      "# s=P(X=0)+P(X=1)\n",
      "s=0;\n",
      "for t in range(0,1):\n",
      "    f=(math.factorial(n))/((math.factorial(t))*(math.factorial(n-t)));\n",
      "    s=s+(f*(pe**t)*((1-pe)**(n-t)));\n",
      "x=1-s;\n",
      "print'Probabilty of more than one error in 10 recieved digits:',x\n",
      "\n",
      "# (b)Using Poisson approximation\n",
      "# P(X=k)~[{(%exp)**(-n*p)}*{((n*p)**k)}]/k factorial\n",
      "s1=0;\n",
      "for k in range(0,1):\n",
      "    j=math.factorial(k);\n",
      "    s1=s1+((math.exp(-n*pe))*(((n*pe)**k)))/j;\n",
      "x1=1.-s1;\n",
      "print'Using Poisson Approximation:',x1"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Probabilty of more than one error in 10 recieved digits: 0.0956179249912\n",
        "Using Poisson Approximation: 0.095162581964\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E23 : Pg 6.23"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Page Number: 6.23\n",
      "# Example 6.23\n",
      "# We find, k=1/(b-a)\n",
      "# (b) if a=1 and b=2,P(|x|<c) where c=1/2\n",
      "a=-1.;\n",
      "b=2.;\n",
      "c=1./2.;\n",
      "\n",
      "k=1./(b-a);\n",
      "# P(|x|<c) = P(-c<=x<=c)\n",
      "# Let y\n",
      "x0=-c;x1=c;\n",
      "y=1.;#integrate('1','x',x0,x1);\n",
      "y1=k*y;\n",
      "print'P(|x|<c):',y1"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "P(|x|<c): 0.333333333333\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E41 : Pg 6.34"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Page Number: 6.34\n",
      "# Example 6.41\n",
      "# Given\n",
      "import math \n",
      "n=16.; # binary digits\n",
      "p=0.01; # Probabilty error due to noise\n",
      "\n",
      "# (a) Average errors per block\n",
      "# E(X)=n*p\n",
      "EofX=n*p;\n",
      "print'Average errors per block:',EofX\n",
      "\n",
      "# (b)Varience of errors of block\n",
      "# s=n*p*(1-p)\n",
      "s=n*p*(1.-p);\n",
      "print'Varience of errors per block:',s\n",
      "\n",
      "\n",
      "# (c) Probability that number of errors per block is bgeater or equal than 4\n",
      "i=4.;\n",
      "# AsP(X>=4)=1=P(X<=3)\n",
      "P3=0;\n",
      "for k in range(0,3):\n",
      "    f=(math.factorial(n))/((math.factorial(k))*(math.factorial(n-k)));\n",
      "    P3=P3+(f*(p**k)*((1-p)**(n-k)));\n",
      "P4=1.-P3;\n",
      "print'Probability that number of errors per block is bgeater or equal than 4:',P4"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Average errors per block: 0.16\n",
        "Varience of errors per block: 0.1584\n",
        "Probability that number of errors per block is bgeater or equal than 4: 0.000507942409291\n"
       ]
      }
     ],
     "prompt_number": 7
    }
   ],
   "metadata": {}
  }
 ]
}