Probability: Random Sample Problem I need some help on the following problem:
Let $X_1$ and $X_2$ be random sample from the pdf
\begin{equation}
f(x) = \begin{cases}
4x^3,&0<x<1\\
0, & \text{otherwise}
\end{cases}
\end{equation}
Obtain $P(X_1X_2\geq 1/4)$.
So here is what I did:
$P(X_1X_2\geq 1/4)=1-P(X_1X_2< 1/4)$
Next, I need to obtain the distribution of $X_1X_2$ in order to evaluate the above probability, but how? I really appreciate if you could give me a solution on this.

You don't need to obtain the distribution of the product $X_1X_2$ to solve the problem.
The probability is determined by integrating the joint density function over the slice of the unit square above the hyperbola $xy=1/4$.
$$\begin{align}
\mathsf P(X_1X_2 \geq \tfrac 1 4)
 & = \int_{1/4}^1 f(x)\;\mathsf P(X_2 \geq \tfrac1{4x}\mid X_1=x)\, \operatorname d x
\\[1ex]
 & = \int_{1/4}^1 \int_{1/4x}^1 f(x)f(y)\,\operatorname d y\operatorname d x
\\[1ex]
 & = 16 \int_{1/4}^1 \int_{1/4x}^1 x^3\; y^3\,\operatorname d y\operatorname d x
\\[1ex]
 & =
\end{align}$$
Can you take it from here?
