The problem presented involves calculating the probability $P[(X, Y) \in A]$, where $A$ is the region defined by $0 < x < 1$ and $\frac{1}{4} < y < \frac{1}{2}$. The function $f(x, y)$ given is a probability density function (PDF) defined as:

$f(x, y) = \frac{x(1 + 3y^2)}{4}, \quad 0 < x < 2, \, 0 < y < 1$ and $f(x, y) = 0$ for values outside of this region.

To solve for $P[(X, Y) \in A]$, we use the following steps:

1. Set up the integral for the probability: Since the PDF $f(x, y)$ is valid within the given bounds, the probability over region $A$ is the double integral of $f(x, y)$ over the bounds defined by $A$:

   $P[(X, Y) \in A] = \int_0^1 \int_{\frac{1}{4}}^{\frac{1}{2}} \frac{x(1 + 3y^2)}{4} \, dy \, dx$

2. Solve the inner integral with respect to $y$:

   $\int_{\frac{1}{4}}^{\frac{1}{2}} \frac{1 + 3y^2}{4} \, dy$ Split the integral:

   = \frac{1}{4} \left[ \int_{\frac{1}{4}}^{\frac{1}{2}} 1 \, dy + 3 \int_{\frac{1}{4}}^{\frac{1}{2}} y^2 \, dy \right]$$ For the first integral: $$\int_{\frac{1}{4}}^{\frac{1}{2}} 1 \, dy = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$ For the second integral: $$\int_{\frac{1}{4}}^{\frac{1}{2}} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{\frac{1}{4}}^{\frac{1}{2}} = \frac{1}{24} - \frac{1}{192} = \frac{7}{192}$$ Thus, the total inner integral becomes: $$\frac{1}{4} \left( \frac{1}{4} + 3 \times \frac{7}{192} \right) = \frac{1}{16} + \frac{21}{768} = \frac{69}{768} = \frac{23}{256}$$

3. Solve the outer integral with respect to $x$: Now, we integrate with respect to $x$:

   $\int_0^1 x \cdot \frac{23}{256} \, dx = \frac{23}{256} \int_0^1 x \, dx = \frac{23}{256} \times \frac{x^2}{2} \bigg|_0^1 = \frac{23}{256} \times \frac{1}{2} = \frac{23}{512}$

Thus, the final probability is:

$P[(X, Y) \in A] = \frac{23}{512}$

Would you like further clarification or details on this solution?

Here are 5 related questions to further expand this topic:

1. How would the probability change if the bounds on $x$ were expanded to $0 < x < 2$?
2. What is the total probability over the entire region $0 < x < 2$, $0 < y < 1$?
3. How can we verify that $f(x, y)$ is a valid probability density function?
4. How would the integral change if the region $A$ were instead defined as $\frac{1}{2} < y < 1$?
5. How does the shape of the function $f(x, y)$ affect the distribution of $X$ and $Y$?

Tip: In double integrals, it's important to carefully choose the order of integration based on the given limits for each variable.