{
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"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
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{
"name": "stdout",
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"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"
]
}
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