A circuit has three inputs A, B and C. Gate 1 is an AND gate with inputs A and B…

Step 1. The expression. $Y = A \cdot B + \overline{C}$.

Step 2. When is $Y = 0$? An OR gate gives $0$ only when both inputs are $0$. So we need $\overline{C} = 0$, which means $C = 1$, and $A \cdot B = 0$, which means that $A$ and $B$ are not both $1$. With $C = 1$ there are four patterns of $A$ and $B$, and one of them, $1, 1$, gives $A \cdot B = 1$. So three patterns give $Y = 0$: $(A, B, C) = (0, 0, 1)$, $(0, 1, 1)$ and $(1, 0, 1)$.

Step 3. When is $Y = 1$? The other $8 - 3 = 5$ patterns.