To solve the given conditional probability problem, we need to calculate $P(1/4 < X < 1/2 | Y = 3/4)$. The conditional probability density function is given as:

$f_{X|Y}(x|y) = \frac{3(x^2 + y^2)}{3y^2 + 1}, \quad 0 < x < 1, \, 0 < y < 1.$

Given $Y = 3/4$, substitute $y = 3/4$ into $f_{X|Y}(x|y)$:

$f_{X|Y}(x|3/4) = \frac{3(x^2 + (3/4)^2)}{3(3/4)^2 + 1}.$

First, calculate the denominator: $3(3/4)^2 + 1 = 3 \cdot \frac{9}{16} + 1 = \frac{27}{16} + \frac{16}{16} = \frac{43}{16}.$

Thus: $f_{X|Y}(x|3/4) = \frac{3(x^2 + 9/16)}{43/16} = \frac{48(x^2 + 9/16)}{43}.$

Now, compute the probability $P(1/4 < X < 1/2 | Y = 3/4)$ by integrating $f_{X|Y}(x|3/4)$ over $x$ from $1/4$ to $1/2$: $P(1/4 < X < 1/2 | Y = 3/4) = \int_{1/4}^{1/2} f_{X|Y}(x|3/4) \, dx.$

Substitute $f_{X|Y}(x|3/4)$ into the integral: $P(1/4 < X < 1/2 | Y = 3/4) = \int_{1/4}^{1/2} \frac{48(x^2 + 9/16)}{43} \, dx.$

Simplify the integral: $P(1/4 < X < 1/2 | Y = 3/4) = \frac{48}{43} \int_{1/4}^{1/2} (x^2 + 9/16) \, dx.$

Break it into two parts: $P(1/4 < X < 1/2 | Y = 3/4) = \frac{48}{43} \left[ \int_{1/4}^{1/2} x^2 \, dx + \int_{1/4}^{1/2} \frac{9}{16} \, dx \right].$

1. Compute $\int_{1/4}^{1/2} x^2 \, dx$: $\int x^2 \, dx = \frac{x^3}{3}.$ Evaluate from $1/4$ to $1/2$: $\left[ \frac{x^3}{3} \right]_{1/4}^{1/2} = \frac{(1/2)^3}{3} - \frac{(1/4)^3}{3} = \frac{1/24} - \frac{1/192} = \frac{8}{192} - \frac{1}{192} = \frac{7}{192}.$

2. Compute $\int_{1/4}^{1/2} \frac{9}{16} \, dx$: $\int \frac{9}{16} \, dx = \frac{9}{16} x.$ Evaluate from $1/4$ to $1/2$: $\left[ \frac{9}{16} x \right]_{1/4}^{1/2} = \frac{9}{16} \cdot \frac{1}{2} - \frac{9}{16} \cdot \frac{1}{4} = \frac{9}{32} - \frac{9}{64} = \frac{18}{64} - \frac{9}{64} = \frac{9}{64}.$

Combine the results: $P(1/4 < X < 1/2 | Y = 3/4) = \frac{48}{43} \left( \frac{7}{192} + \frac{9}{64} \right).$

Simplify the terms inside the parentheses: $\frac{7}{192} + \frac{9}{64} = \frac{7}{192} + \frac{27}{192} = \frac{34}{192} = \frac{17}{96}.$

Thus: $P(1/4 < X < 1/2 | Y = 3/4) = \frac{48}{43} \cdot \frac{17}{96} = \frac{816}{4128} = \frac{17}{86}.$

The final probability is: $P(1/4 < X < 1/2 | Y = 3/4) = \frac{17}{86}.$

Let me know if you need further clarification or any additional details! Here are 5 questions to expand on this:

1. What is a conditional probability density function and how does it differ from a joint PDF?
2. How do you verify whether a given PDF is valid?
3. Can you explain how integration applies to finding probabilities in continuous random variables?
4. What are the properties of $f_{X|Y}(x|y)$ in terms of its limits and normality conditions?
5. How would the calculation change if the bounds of $X$ or $Y$ were modified?

Tip: Always double-check the bounds and normalization when solving probability density functions to ensure accurate results.