Question 1164365: The random variables (X,Y) have joint pdf f(x,y) = 1/32 for the region 0 < x < 8, 0 < y < 4. The pdf is zero outside this region.
iii. Find the probability P(X+Y < 4).
Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=2
import scipy.integrate as integrate
# Define the pdf
def f(x, y):
if 0 < x < 8 and 0 < y < 4:
return 1/32
return 0
# Define the probability condition for the integral
# P(X + Y < 4)
# Bounds for y: 0 to min(4, 4 - x) -> but 4-x <= 4 for x >= 0
# Bounds for x: 0 to 4 (since 4-x must be > 0)
prob, error = integrate.dblquad(lambda y, x: 1/32, 0, 4, lambda x: 0, lambda x: 4 - x)
print(f"Probability: {prob}")
print(f"Fraction: {prob * 32}/32")
```
```text?code_stdout&code_event_index=2
Probability: 0.25
Fraction: 8.0/32
```
To find the probability $P(X + Y < 4)$ for the given joint probability density function, we can use either double integration or geometric area calculations since the distribution is uniform.
### **1. Identify the Region and PDF**
The joint pdf is given by:
$$f(x, y) = \frac{1}{32} \quad \text{for } 0 < x < 8, \ 0 < y < 4$$
The total area of the sample space is $8 \times 4 = 32$, and the total volume (probability) is $32 \times \frac{1}{32} = 1$.
### **2. Define the Event Region**
We want to find the probability of the event $X + Y < 4$, which can be rewritten as $Y < 4 - X$.
Additionally, the variables are constrained by the original region:
* $x > 0$
* $y > 0$
* $x < 8$
* $y < 4$
Combining $y < 4 - x$ with $y > 0$, we find that $4 - x > 0$, which implies **$x < 4$**.
Thus, the region of interest is a triangle with vertices at $(0,0)$, $(4,0)$, and $(0,4)$.
### **3. Calculation (Geometric Method)**
Since $f(x, y)$ is a constant (uniform distribution), the probability is simply the area of the event region multiplied by the value of the pdf:
$$\text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8$$
$$P(X + Y < 4) = \text{Area} \times f(x, y) = 8 \times \frac{1}{32} = \frac{8}{32} = \frac{1}{4}$$
### **4. Calculation (Integration Method)**
We can verify this by integrating the joint pdf over the triangular region:
$$P(X + Y < 4) = \int_{0}^{4} \int_{0}^{4-x} \frac{1}{32} \, dy \, dx$$
Evaluating the inner integral:
$$\int_{0}^{4-x} \frac{1}{32} \, dy = \left[ \frac{y}{32} \right]_{0}^{4-x} = \frac{4-x}{32}$$
Evaluating the outer integral:
$$\int_{0}^{4} \frac{4-x}{32} \, dx = \frac{1}{32} \left[ 4x - \frac{x^2}{2} \right]_{0}^{4} = \frac{1}{32} \left( 16 - \frac{16}{2} \right) = \frac{8}{32} = \frac{1}{4}$$
**Final Result:**
$$P(X + Y < 4) = 0.25$$
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