Let X and Y be two statistically independent random variables uniformly distributed in the range (−1, 1) and (−2, 1) respectively. Let Z = X +Y , then the probability that Z<=-2

To solve the problem of finding the probability that \( Z = X + Y \leq -2 \), where \( X \) and \( Y \) are statistically independent random variables uniformly distributed in the ranges \( (-1, 1) \) and \( (-2, 1) \) respectively, let's go step by step.

### 1. Definitions and Ranges
- \( X \) is uniformly distributed over \( (-1, 1) \).
  \[
  f_X(x) = \frac{1}{2}, \quad -1 < x < 1
  \]
- \( Y \) is uniformly distributed over \( (-2, 1) \).
  \[
  f_Y(y) = \frac{1}{3}, \quad -2 < y < 1
  \]

### 2. Joint Distribution
Since \( X \) and \( Y \) are independent, the joint probability density function (PDF) of \( X \) and \( Y \) is:
\[
f_{X,Y}(x, y) = f_X(x) \cdot f_Y(y) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}, \quad -1 < x < 1, \ -2 < y < 1
\]

### 3. Probability Calculation
We want to find the probability \( P(Z \leq -2) \), where \( Z = X + Y \).

Let's consider the inequality:
\[
Z \leq -2 \quad \text{which implies} \quad X + Y \leq -2
\]

Given the ranges of \( X \) and \( Y \):
- \( X \) ranges from \(-1\) to \(1\).
- \( Y \) ranges from \(-2\) to \(1\).

For \( Z = X + Y \leq -2 \), we can express \( Y \) in terms of \( X \):
\[
Y \leq -2 - X
\]

The probability can be computed by integrating the joint PDF over the region defined by the above inequality:

### 4. Integration Boundaries
1. **For \( X = -1 \):**
   \[
   Y \leq -2 - (-1) = -1
   \]
   Since \( Y \) must be between \(-2\) and \(1\), this condition is valid.

2. **For \( X = 1 \):**
   \[
   Y \leq -2 - 1 = -3
   \]
   Since \( Y \) is bounded between \(-2\) and \(1\), this condition is never satisfied.

The region of integration is:
\[
\int_{-1}^{0} \int_{-2}^{-2-x} \frac{1}{6} \, dy \, dx
\]

### 5. Perform the Integration
First, integrate with respect to \( Y \):
\[
\int_{-2}^{-2-x} \frac{1}{6} \, dy = \frac{1}{6} \left[ y \right]_{-2}^{-2-x} = \frac{1}{6}(-2-x + 2) = \frac{-x}{6}
\]

Now, integrate with respect to \( X \):
\[
\int_{-1}^{0} \frac{-x}{6} \, dx = \frac{-1}{6} \int_{-1}^{0} x \, dx = \frac{-1}{6} \left[ \frac{x^2}{2} \right]_{-1}^{0}
\]
\[
= \frac{-1}{6} \left( 0 - \frac{1}{2} \right) = \frac{1}{12}
\]

### 6. Final Probability
Thus, the probability that \( Z \leq -2 \) is:
\[
P(Z \leq -2) = \frac{1}{12}
\]

### Conclusion
The probability that \( Z \leq -2 \) is \( \frac{1}{12} \).

Would you like any more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the probability that \( Z \) equals a specific value, say \( Z = 0 \)?
2. How does the range of \( Z = X + Y \) change if the distribution of \( X \) or \( Y \) changes?
3. What is the expected value of \( Z = X + Y \)?
4. How do you compute the probability \( Z \) is within a certain range, say \( -1 \leq Z \leq 1 \)?
5. What is the variance of \( Z = X + Y \)?

**Tip:** For independent random variables, the sum of their means gives the mean of their sum, and the sum of their variances gives the variance of their sum.

Let X and Y be two statistically independent random variables uniformly distributed in the range (−1, 1) and (−2, 1) respectively. Let Z = X + Y , then the probability that Z<=-2

Explore how to calculate the probability that Z = X + Y is less than or equal to -2 when X and Y are independent random variables uniformly distributed over (−1, 1) and (−2, 1). The solution includes step-by-step integration, relevant probability concepts, and key formulas.

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