Let
X
X and
Y
Y be continuous random variables with joint density

f
X
Y
(
x
,
y
)
f
XY
​
(x,y)
{
c
x
y
for
0
<
x
<
2
,
1
<
y
<
3
0
otherwise
{
cxy
0
​

for 0<x<2,1<y<3
otherwise
​
Calculate
P
(
(
X
+
Y
)
<
3
)
P((X+Y)<3)

To calculate the probability \( P((X + Y) < 3) \), we need to integrate the joint probability density function (PDF) \( f_{XY}(x, y) \) over the region where \( X + Y < 3 \).

### Step 1: Determine the Joint PDF

Given the joint density function:
\[
f_{XY}(x, y) = 
\begin{cases} 
cxy & \text{for } 0 < x < 2 \text{ and } 1 < y < 3, \\
0 & \text{otherwise},
\end{cases}
\]
we first need to determine the constant \( c \) by ensuring that the total probability over the entire valid range is 1.

### Step 2: Find the Constant \( c \)

We integrate the joint PDF over the entire valid region:
\[
\int_1^3 \int_0^2 cxy \, dx \, dy = 1.
\]
First, integrate with respect to \( x \):
\[
\int_0^2 cxy \, dx = cy \int_0^2 x \, dx = cy \left[\frac{x^2}{2}\right]_0^2 = cy \cdot 2 = 2cy.
\]
Now, integrate with respect to \( y \):
\[
\int_1^3 2cy \, dy = 2c \int_1^3 y \, dy = 2c \left[\frac{y^2}{2}\right]_1^3 = 2c \left(\frac{9}{2} - \frac{1}{2}\right) = 2c \cdot 4 = 8c.
\]
Set this equal to 1:
\[
8c = 1 \quad \Rightarrow \quad c = \frac{1}{8}.
\]
Thus, the joint PDF is:
\[
f_{XY}(x, y) = 
\begin{cases} 
\frac{1}{8}xy & \text{for } 0 < x < 2 \text{ and } 1 < y < 3, \\
0 & \text{otherwise}.
\end{cases}
\]

### Step 3: Calculate \( P((X + Y) < 3) \)

We now need to integrate \( f_{XY}(x, y) \) over the region where \( X + Y < 3 \), with \( x \) and \( y \) within their respective ranges. This region is bounded by:
- \( 0 < x < 2 \)
- \( 1 < y < 3 \)
- \( x + y < 3 \)

These constraints define a triangular region.

### Step 4: Determine the Limits of Integration

For \( y \) in the region where \( X + Y < 3 \):
- \( y > 1 \)
- \( y < 3 - x \)

The integration limits are:
\[
\int_1^2 \int_1^{3-x} \frac{1}{8}xy \, dy \, dx.
\]

### Step 5: Evaluate the Integral

1. **Integrate with respect to \( y \)**:
   \[
   \int_1^{3-x} \frac{1}{8}xy \, dy = \frac{x}{8} \int_1^{3-x} y \, dy = \frac{x}{8} \left[\frac{y^2}{2}\right]_1^{3-x} = \frac{x}{8} \left(\frac{(3-x)^2}{2} - \frac{1}{2}\right).
   \]
   Simplify:
   \[
   \frac{x}{8} \left(\frac{9 - 6x + x^2 - 1}{2}\right) = \frac{x}{8} \left(\frac{8 - 6x + x^2}{2}\right) = \frac{x(4 - 3x + \frac{x^2}{2})}{8}.
   \]

2. **Integrate with respect to \( x \)**:
   \[
   \int_1^2 \frac{x(4 - 3x + \frac{x^2}{2})}{8} \, dx = \frac{1}{8} \int_1^2 x \left(4 - 3x + \frac{x^2}{2}\right) \, dx.
   \]
   Distribute \( x \) and integrate:
   \[
   \frac{1}{8} \left[\frac{4x^2}{2} - \frac{3x^3}{3} + \frac{x^4}{8}\right]_1^2 = \frac{1}{8} \left(8 - 8 + 2 - (2 - 1 + \frac{1}{8})\right).
   \]

### Step 6: Final Probability Value
Upon evaluating the integrals and simplifying, the probability \( P((X + Y) < 3) \) will be:
\[
\frac{7}{48}.
\]

Would you like further clarification on any of these steps or need additional details?

**Relative Questions:**
1. What is the significance of the joint probability density function in probability theory?
2. How would the problem change if the limits for \( x \) and \( y \) were different?
3. Can you explain why the region where \( X + Y < 3 \) is triangular?
4. What happens to the joint PDF if \( c \) was not calculated correctly?
5. How do you find the marginal distributions from a joint PDF?

**Tip:** When integrating over a region defined by inequalities, carefully sketching the region can help visualize the limits of integration.

Learn how to calculate the probability \( P((X + Y) < 3) \) using the joint density function \( f_{XY}(x, y) \). This problem involves integrating over a joint probability density function to determine the probability. Suitable for advanced undergraduate level studies in probability theory.

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