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  1. Prepare
  2. Python
  3. Itertools
  4. Iterables and Iterators
  5. Discussions

Iterables and Iterators

  • Solspin
     + 0 comments

    I know this is for itertools practice, but without itertools the answer is quite clean. Using 1 minus the multiplicative probability of not choosing 'a'.

    N,s,n = int(input()),input(),int(input())
x = 1
b = N-s.count('a')
for i in range(n):
    x = x*((b-i)/(N-i)) 
print(1-x)