What is the Probability of ‘begining’ given ‘the’?

Question

Using an NLTK Conditional Frequency Distribution and the nltk.bigrams function, train a bigram model on the Genesis:

text = nltk.corpus.genesis.words('english-kjv.txt')
bigrams = nltk.bigrams(text)
cfd = nltk.ConditionalFreqDist(bigrams)
Answer the following questions

What is the Probability of ‘begining’ given ‘the’?
What is the probability of ‘the’?

Note: The probabilities you give as an answer MUST be probabilities computable from this corpus.

Hi, can some help me? this is in the nltk book. When I got it, I got 78% which does not make sense. Im trying to compute this in Python.

Answer 1

There is sort of a difference between probability of 'beginning' intersect 'the'

p('beginning','the')

and probability of 'beginning' given 'the':

p('beginning'|'the') = p('beginning','the') / p('the')

try:

from collections import Counter

import nltk

text = nltk.corpus.genesis.words('english-kjv.txt')
bigrams = nltk.bigrams(text)
cfd_bigrams = Counter(bigrams)
cfd_unigrams = Counter(list(text))

print "p('said','unto') =", cfd_bigrams[u'said', u'unto'] / float(sum(cfd_bigrams.values()))

print "p('said'|'unto') =", (cfd_bigrams[u'said', u'unto'] / float(sum(cfd_bigrams.values()))) / cfd_unigrams[u'unto']

print "p('beginning','the') =", cfd_bigrams[u'beginning', u'the']

[out]:

p('said','unto') = 0.00397649844738
p('said'|'unto') = 6.73982787691e-06
p('beginning','the') = 0