P(A,B,C,D,E,F) = P(F)P(E|B,D,F)P(D|A)P(C|B)P(B|A)P(A)

The parent-less nodes (A and F) only require a single probability, since eg if we store P(A=yes), we can work out P(A=no) as 1-P(A=yes). B,C and D all require 2 conditional probabilities. For example, for B we need P(B=yes|A=yes) and P(B=yes|A=no). We can work out P(B=no|A=yes) and P(B=no|A=no) by subtracting from 1 as before. Finally E requires 8 conditional probabilities

In total we need 1+2+2+2+8+1 = 16 probabilities.

If we could not assume the given Bayes net structure, we would need 2^6 - 1 2*2*2*2*2*2-1 = 63 probabilities.