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 A  sample_notebooks/ArshadKhan/Chap10.ipynb

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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter - 10 : Introduction to theory of probability"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## page no 437 Ex 10.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "probability of each outcome=0.12\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "# referred to fig 10.1 on the page no. 435\n",
+    "# the occurance of each outcome is essumed to be equal.\n",
+    "n=8 # number of outcomes\n",
+    "p=1/n#\n",
+    "print \"probability of each outcome=%0.2f\"%p"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## page no 438 Ex no 10.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "probability = 0.1667\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "m=6 # enter the number of faces \n",
+    "n=2# enter the number of dice \n",
+    "l=m**n ## total number of outcomes = 36\n",
+    "a=7 #\"enter the number which is to be obtained as the sum of dice\n",
+    "c=0 # # counter value for favorable outcome\n",
+    "for i in range(1,7):\n",
+    "    for j in range(1,7):\n",
+    "        if (i+j==a):\n",
+    "            c=c+1#\n",
+    "        else :\n",
+    "            continue\n",
+    "   \n",
+    "  \n",
+    "\n",
+    "p=c/l#\n",
+    "print \"probability = %0.4f\"%p\n",
+    "  "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## page no 438 Ex no 10.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "probability = 0.4075\n"
+     ]
+    }
+   ],
+   "source": [
+    "from numpy.random import gamma\n",
+    "from __future__ import division\n",
+    "m=2# the number of faces \n",
+    "n=4 #the number of tosses\n",
+    "l=m**n ## l is total number of outcomes = 16\n",
+    "a=2 #exact no of heads\n",
+    "p=gamma (n+1)/(gamma(n+1-a) * gamma (a+1))## to find combination\n",
+    "print \"probability = %0.4f\"%(p/l)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## page no 440 Ex no 10.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "total probability = 0.0023\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "a=52# # total no of cards in a deck\n",
+    "b=6# the no of cards to be drawn\n",
+    "pA1= b/a## probability of getting first red ace =pA1\n",
+    "#the cards are drawn in succession without replacement, therefore the probability that the 2nd card will be the red ace = pA2\n",
+    "pA2=1/(a-1)#\n",
+    "p= pA1*pA2\n",
+    "print \"total probability = %0.4f\"%p"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## page no 441 Ex no 10.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "probability = 1.04e-10\n"
+     ]
+    }
+   ],
+   "source": [
+    "from numpy.random import gamma\n",
+    "from __future__ import division\n",
+    "\n",
+    "# This problem is based on Bernoulli Trials formula which is P( k successes in n trials ) = n!*p**k *(1-p)**(n-k)/k!*(n-k)!22\n",
+    "# hence the probability of finding 2 digits wrong in a sequence of 8 digits is\n",
+    "\n",
+    "k=2 # no. of successes\n",
+    "p= 1e-5# probability of success\n",
+    "n=8 #\"no. of trials\n",
+    "A=gamma (n+1)* (p**k)*((1-p)**(n-k))/(gamma(k)*gamma(n+1-k))#\n",
+    "print \"probability = %0.2e\"%A"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## page no 446 Ex no 10.9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "probability =  0.0231481481481\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "m=6 # enter the number of faces\n",
+    "n=3 # enter the number of dice\n",
+    "l=m**n ## l is total number of outcomes \n",
+    "a=8# the number which is to be obtained as the sum of dice\n",
+    "c=0 # # counter value for favorable outcome\n",
+    "for i in range(1,7):\n",
+    "    for j in range(1,7):\n",
+    "        if(i+j==a):\n",
+    "            c=c+1\n",
+    "        else :\n",
+    "            continue\n",
+    "   \n",
+    "p=c/l#\n",
+    "print \"probability = \",p\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## page no 447 Ex no 10.10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Py(1) =  0.59235668\n",
+      "Py0= 0.40764332\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "Pe=0.6898 # error probability\n",
+    "Q= 0.2567 # the probability of transmitting 1 #Hence probability of transmitting zero is 1-Q = P\n",
+    "P=1-Q#\n",
+    "Px_1=Q#\n",
+    "Px0=P#\n",
+    "# If x and y are transmitted digit and received digit then for BSC P(y=0/x=1) = P(y=1/x=0) = Pe , P(y=0/x=10) = P(y=1/x=1) = 1-Pe\n",
+    "# to find the probability of receiving 1 is Py(1) = Px(0)*P(y=1/x=0) + Px(1)*P(y=1/x=1)\n",
+    "Py_1= ((1-Q)* Pe) + (Q *(1-Pe))#\n",
+    "print \"Py(1) = \",Py_1\n",
+    "Py0=((1-Q)*(1-Pe)) + (Q*Pe)\n",
+    "print \"Py0=\",Py0"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## page no 448 Ex 10.11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "probability of error =  6.04e-05\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "Px0=.4\n",
+    "Px1=.6 \n",
+    "PE0=10**-6 \n",
+    "PE1=10**-4 ## given\n",
+    "PE=(Px0*PE0) + (Px1*PE1)# formula for probability of error\n",
+    "print \"probability of error = \",PE"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## page no 472 Ex no 10.20"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "width or spread of Gaussian PDF = 0.985977359605\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "from math import pi,sqrt,exp\n",
+    "#Gaussian PDF: Q(x)= %e**((-x**2)/2)/ (x*sqrt(2*pi))\n",
+    "x=2.5 #  input for the function Q \n",
+    "Q_x = (exp(-(x**2)/2))/ (x*sqrt(2*pi))\n",
+    "P=1-(2*Q_x)#\n",
+    "print \"width or spread of Gaussian PDF =\",P\n",
+    "## P gives the width or spread of Gaussian PDF"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## page no 479 Ex no 10.21"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "SNR improvement = 12.90 dB\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "from math import log\n",
+    "# formula for estimate error E is E = mk** - mk = a1* mk-1 +a2* mk-2 -mk\n",
+    "#given: various values of correlation (mk*mk)'= (m**2)',(mk*mk-1)'= .825* (m**2)',(mk*mk-2)'= .562*(m**2)',(mk*mk-3)'= .825*(m**2)' , R02=.562(m**2)', a1=1.1314, a2= -0.3714\n",
+    "# mean square error is given by I=(E**2)'=[1-((.825*a1)+(.562*a2))]*(m**2)'= .2753*(m**2)'\n",
+    "\n",
+    "m=1#\n",
+    "I=.2753*(m**2)#\n",
+    "S=10*log ((m**2)/I)#\n",
+    "print \"SNR improvement = %0.2f dB\"%S"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}