Refer to the logic circuit given below. For two inputs (A=1, B=1)...

From the given logic circuit, we can determine the Boolean expression for the output $Y$.

The output of the NOT gate is $\bar{A}$.

The inputs to the OR gate are $\bar{A}$ and $B$, so its output is $\bar{A} + B$.

The inputs to the AND gate are $\bar{A}$ and $B$, so its output is $\bar{A} \cdot B$.

These two outputs are fed into a NOR gate. The final output $Y$ is given by:

$Y = \overline{(\bar{A} + B) + (\bar{A} \cdot B)}$

Simplifying the Boolean expression:

$Y = \overline{\bar{A} + B + \bar{A}B}$

$Y = \overline{\bar{A} + B(1 + \bar{A})}$

$Y = \overline{\bar{A} + B}$

Using De Morgan's laws:

$Y = \overline{\bar{A}} \cdot \bar{B} = A \cdot \bar{B}$

Now, we evaluate $Y$ for the given input combinations:

For $A = 1, B = 1$:

$Y = 1 \cdot \bar{1} = 1 \cdot 0 = 0$

For $A = 0, B = 1$:

$Y = 0 \cdot \bar{1} = 0 \cdot 0 = 0$

Therefore, the outputs are $0$ and $0$ respectively.

Answer: $0, 0$ respectively