Recent content by vlad1234

HOpe I'm not wrong this time . P ( N , L ) = \left\{2/2^{N} , if N = L + 1 \left\{( 2^{ N-L+1} + 2^{N-L-2} + 2 ) / 2^{N} , if N = L + 2 \left\{( 2^{N-L-1} + 2^{N-L-2} + \Sigma^{N-L-2}_{K=1}( 2^{N-L-2-K} * 2^{K} ) + 2^{N-L-1} ) / 2^{N}

Oh , i made a mistake , sorry. Let me fix it . C^{l}_{n} = \left(^{l}_{n}\right)

Hi , i think this is the answer P(n,l) = [SIZE="4"]C^{l}_{n} / 2^{n}