1. **Problem Statement:** We want to find the probability $P(X + Y \leq 5)$ for a continuous bivariate random variable $(X,Y)$ with joint density $g(x,y)$ defined over $0 < X < 4$ and $0 < Y < 4$. 2. **Key Idea:** The event $X + Y \leq 5$ corresponds to the region under the line $y = 5 - x$ within the square $0 < x < 4$, $0 < y < 4$. 3. **Integral Setup:** The probability is the double integral of $g(x,y)$ over the region where $x$ and $y$ satisfy $0 < x < 4$, $0 < y < 4$, and $y \leq 5 - x$. 4. **Check each option:** - (a) $\int_0^4 \int_0^{5-x} g(x,y) \, dy \, dx$ integrates over $x$ from 0 to 4 and $y$ from 0 to $5-x$, which exactly covers the region $X+Y \leq 5$ inside the square. **Correct**. - (b) $\int_0^1 \int_0^{5-x} g(x,y) \, dy \, dx$ only integrates $x$ from 0 to 1, missing the part where $1 < x < 4$. **Incorrect**. - (c) $\int_0^1 \int_0^4 g(x,y) \, dy \, dx + \int_1^4 \int_0^{5-x} g(x,y) \, dy \, dx$ splits the region: for $x$ in $[0,1]$, $y$ goes up to 4 (since $5-x \geq 4$), and for $x$ in $[1,4]$, $y$ goes up to $5-x$. This covers the entire region $X+Y \leq 5$ inside the square. **Correct**. - (d) $1 - \int_1^4 \int_{5-x}^4 g(x,y) \, dy \, dx$ subtracts the probability of the region where $x$ is from 1 to 4 and $y$ is above $5-x$ (outside the event) from 1. This correctly gives $P(X+Y \leq 5)$. **Correct**. - (e) $1 - \int_0^1 \int_{5-x}^4 g(x,y) \, dy \, dx$ subtracts the region above $5-x$ for $x$ in $[0,1]$ only, missing the part for $x$ in $[1,4]$. **Incorrect**. - (f) $1 - \int_0^1 \int_0^4 g(x,y) \, dy \, dx - \int_1^4 \int_{5-x}^4 g(x,y) \, dy \, dx$ subtracts the entire probability for $x$ in $[0,1]$ plus the region above $5-x$ for $x$ in $[1,4]$. This removes too much and does not equal $P(X+Y \leq 5)$. **Incorrect**. **Final correct integrals:** (a), (c), and (d).