text: Q:
  
  Probabilities, Unigram and Bigram
  
  Assume that we have these bigram and unigram data:( Note: not a real data)
  bigram:
  #a(start with a) =21
  bc= 42
  cf= 32
  de= 64
  e#= 23     
  unigram:
  # 43
  a= 84
  b=123
  c=142
  f=161
  d=150
  e=170
  what is the probability of generating a word like "abcfde"?
  I think for having a word starts with a the probability is 21/43. How about bc? is it like bc/b?
  
  A:
  
  Augment the string "abcde" with # as start and end markers to get #abcde#. Now, as @Yuval Filmus pointed out,  we need to make some assumption about the kind of model that generates this data. Because we have both unigram and bigram counts, we can assume a bigram model. In a bigram (character) model, we find the probability of a word by multiplying conditional probabilities of successive pairs of characters, so:
  $\Pr[\#abcde\#] = \Pr(a|\#)*\Pr(b|a)*\Pr(c|b)*\Pr(d|c)*\Pr(e|d)*\Pr(\#|e) $
  To find the conditional probability of a character $c_2$ given its preceding character $c_1$, $\Pr(c_2|c_1)$, we divide the number of occurrences of the bigram $c_1c_2$ by the number of occurrences of the unigram $c_1$.
  So, for example $\Pr(e|d) = count(de)/count(d) = 64/150$
  
  