Refer to the logic circuit given below. For two inputs \((A=1, B=1)\) and \((A=0, B=1)\), output \((Y)\) will be __________.

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