From where are the following probabilities?

We consider BSCε with ε = 0,1 and block code C = {c1, c2} with the code words c1 = 010 and c2 = 101. On the received word y we use the decoder D = {D1,D2} which decodes the word to the code word which has the lowest hamming distance to y. Determine D1 and D2 and the global error probability ERROR(D) if the code words have the same probability. Hint: To an output y there exists only one x which gets to a failing decoding. (Y = 100 will only gets wrong decoded if we sent message is x = c1 = 010.) So the term (1− p(D(y)|y)) is equal to Pr[X = x|Y = y] for a suitable x.

Nun

$ $ \begin{aligned} &\text { Hamming-Distance: }\ &\begin{array}{c|cc} \text { Code } & 010 & 101 \ \hline \hline 000 & 1 & 2 \ \hline 001 & 2 & 1 \ \hline 010 & 0 & 3 \ \hline 011 & 1 & 2 \ \hline 100 & 2 & 1 \ \hline 101 & 3 & 0 \ \hline 110 & 1 & 2 \ \hline 111 & 2 & 1 \ \hline \end{array} \end{aligned}$ $ $ $ \left.D_{1}=\{000,010,011,110\} \text { (Decides for } 010\right)$ $ \left.D_{2}=\{001,100,101,111\} \text { (Decides for } 101\right)$ $ $ $ \begin{aligned} E R R O R(D) &=\sum_{y \in \Sigma_{A}^{3}} p(y)(1-p(D(y) | y)) \ &=\overbrace{2 \cdot p(y)(1-p(D(y) | y))}+\quad \overbrace{6 \cdot p(y)(1-p(D(y) | y))}^{ } \ &=2 \cdot\left(\frac{729}{2000}+\frac{1}{2000}\right)\left(\frac{7}{250}\right)+6 \cdot\left(\frac{81}{2000}+\frac{9}{2000}\right)\left(\frac{757}{1000}\right) \end{aligned}$ $ How do I get to the probabilities $ $ \frac{7}{250}$ $ and $ $ \frac{757}{1000}$ $ ??

I don’t get this calculation. It should be right. But I don’t get how to get to these probabilities.

Could someone explain me this?