The correct truth table for the following logic circuit is :

• Correct answer
  A

  $\begin{array}{|l|l|l|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \hline \end{array}$
• B

  $\begin{array}{|l|l|l|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \hline \end{array}$
• C

  $\begin{array}{|l|l|l|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \\ \hline \end{array}$
• D

  $\begin{array}{|l|l|l|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \hline \end{array}$

Hence, the correct answer is option: 'A'.