The problem describes two random variables $X$ and $Y$ with a given joint probability density function (PDF) $f(x, y) = 2y$, where $0 < x < 1$ and $0 < y < 1$. The question asks to find the probability:

$P(X^2 < Y < X)$

Step-by-Step Solution:

1. Define the Region: We are interested in the probability that $X^2 < Y < X$. To find this, we need to identify the region of the $(x, y)$-plane where this condition holds, given that $0 < x < 1$ and $0 < y < 1$.

   The condition $X^2 < Y < X$ describes the area between the curves $y = X^2$ and $y = X$.

2. Set up the Probability Expression: The probability can be written as:

   $P(X^2 < Y < X) = \int_0^1 \int_{X^2}^{X} f(x, y) \, dy \, dx$

   Since $f(x, y) = 2y$, we substitute it into the integral:

   $P(X^2 < Y < X) = \int_0^1 \int_{X^2}^{X} 2y \, dy \, dx$

3. Evaluate the Inner Integral: The inner integral with respect to $y$ is:

   $\int_{X^2}^{X} 2y \, dy = \left[ y^2 \right]_{X^2}^{X} = X^2 - (X^4)$

   Simplifying, we get:

   $X^2 - X^4$

4. Evaluate the Outer Integral: Now, integrate with respect to $x$:

   $P(X^2 < Y < X) = \int_0^1 (X^2 - X^4) \, dx$

   Break this into two separate integrals:

   $\int_0^1 X^2 \, dx - \int_0^1 X^4 \, dx$

   These integrals are straightforward:

   $\int_0^1 X^2 \, dx = \frac{1}{3}, \quad \int_0^1 X^4 \, dx = \frac{1}{5}$

   Therefore, the probability is:

   $P(X^2 < Y < X) = \frac{1}{3} - \frac{1}{5} = \frac{5}{15} - \frac{3}{15} = \frac{2}{15}$

So, the correct probability is $\frac{2}{15}$, which matches option (3) in the provided answers.

Would you like further details or clarifications on this? Here are some related questions for practice:

1. How do you generalize the method to find $P(X^a < Y < X^b)$ for arbitrary $a$ and $b$?
2. What would the solution look like if the PDF was a different function, say $f(x, y) = 3xy$?
3. How does changing the limits of $X$ and $Y$ affect the probability calculation?
4. What is the role of the normalization condition in determining the validity of a PDF?
5. How would the solution change if the support of $X$ and $Y$ was from $0$ to $2$ instead of $0$ to $1$?

Tip: When dealing with probability density functions, always ensure the PDF is normalized (i.e., integrates to 1) over the given range to ensure it is valid.