Inside two heads

The probability to get $k$ heads out of $n$ coins is $\frac { \left( \begin{matrix} n \\ k \end{matrix} \right) }{ 2^{ n } } = \frac { n! }{ { 2 }^{ n }(n-k)!k! }$ .

So, the probability to get 2 or less heads for every time you toss the coin is: $\frac { 1 }{ 2^{ n } } \quad \sum _{ k=0 }^{ 2 }{ \left( \begin{matrix} n \\ k \end{matrix} \right) }$ .

Finally, if you toss the coin with $n=4,3,2$ coins, the final probability is: $\prod _{ n=2 }^{ 4 }{ \left( \frac { 1 }{ 2^{ n } } \quad \sum _{ k=0 }^{ 2 }{ \left( \begin{matrix} n \\ k \end{matrix} \right) } \right) } =\frac { 77 }{ 128 }$ .

Hence, the answer is $77+128=\boxed{205}$

Let $X$ be the number of heads turn out. Then for step 1:

$P_1(0\le X\le 2)=P_1(X=0)+P_1(X=1)+P_1(X=2)=\frac{1}{16}+\frac{4}{16}+\frac{6}{16}=\frac{11}{16}$

Similarly, step 2:

$P_2(0\le X\le 2)=\frac{1}{8}+\frac{3}{8}+\frac{3}{8}=\frac{7}{8}$

Step 3:

$P_3(0\le X\le 2)=1$

The three steps together:

$P(0\le X\le 2)=P_1P_2P_3=\frac{11}{16}\times \frac{7}{8}\times 1=\frac{77}{128}=\frac{a}{b}$

$\Rightarrow a=77$ , $b=128$ and $a+b=\boxed{205}$ .

(0.5 x + 0.5 y)^4 = (x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + y^4)/ 16;

(0.5 x + 0.5 y)^3 = (x^3 + 3 x^2 y + 3 x y^2 + y^3)/ 8;

(0.5 x + 0.5 y)^2 = (x^2 + 2 x y + y^2)/ 4;

(11/ 16) else (5/ 16)

(7/ 8) else (1/ 8)

(4/ 4) else (0/ 4)

(11/ 16)(7/ 8)(4/ 4) = 77/ 128 for 205

else 1 - (5/ 16) - (1/ 8) - (0/ 4) = 9/ 16 for 25.

Answer is 205 for each and every steps are successful to achieve (AND Gates).