To solve this problem, we need to analyze the given logic circuit.$\\$The circuit consists of:$\\$• An AND gate with inputs $A$ and $\overline{B}$ (NOT B).$\\$• Another AND gate with inputs $A$ and $B. \\$• An OR gate combining the outputs of the two AND gates.$\\$Let's denote:$\\$• $X = A \cdot \overline{B}$ (output of the first AND gate).$\\$• $Y = A \cdot B$ (output of the second AND gate).$\\$• $C = X + Y$ (output of the OR gate).$\\$Now, construct the truth table:$\\ \begin{array}{c|c|c|c|c|c} A & B & \overline{B} & X & Y & C \\ \hline 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 & 1 \\ \end{array} \\$Explanation:$\\$• When $A = 0,$ both $X$ and $Y$ are 0, so $C = 0. \\$• When $A = 1$ and $B = 0,$ $X = 1$ and $Y = 0,$ so $C = 1. \\$• When $A = 1$ and $B = 1,$ $X = 0$ and $Y = 1,$ so $C = 1. \\$This truth table corresponds to Option 3.