91Ó°ÊÓ

The marginal pdf of \(X_{1}\), denoted as \(f_{X_{1}}(x_{1})\), can be found by integrating the joint pdf over all possible values of \(x_{2}\). In this case, since \(0<x_{1}<x_{2}<1\), the limits for \(x_{2}\) are \(x_{1}\) and 1.\[f_{X_{1}}(x_{1})=\int_{x_{1}}^{1} 15x_{1}^{2} x_{2} dx_{2} = 15x_{1}^{2}\left[\frac{x_{2}^{2}}{2}\right]_{x_{1}}^{1} = 15x_{1}^{2}\left[\frac{1}{2}-\frac{x_{1}^{2}}{2}\right] = 15x_{1}^{2}\left[\frac{-x_{1}^{2}+1}{2}\right]\]Similarly, the marginal pdf of \(X_{2}\), denoted as \(f_{X_{2}}(x_{2})\), can be found by integrating the joint pdf over all possible values of \(x_{1}\). In this case, the limits for \(x_{1}\) are 0 and \(x_{2}\).\[f_{X_{2}}(x_{2})= \int_{0}^{x_{2}} 15x_{1}^{2} x_{2} dx_{1} = x_{2}\left[5x_{1}^{3}\right]_{0}^{x_{2}} = 5x_{2}^{4}\]

This is the probability that \(X_{1}+X_{2}\) is less than or equal to 1. Mathematically, \(X_{1}+X_{2} \leq 1\) implies \(X_{1} \leq 1-X_{2}\). The region of integration is between 0 and \(\frac{1}{2}\) for both \(X_{1}\) and \(X_{2}\). \[P\left(X_{1}+X_{2} \leq 1\right) = \int_{0}^{\frac{1}{2}} \int_{0}^{1-x_{2}} 15x_{1}^{2} x_{2} dx_{1} dx_{2} = \int_{0}^{\frac{1}{2}} x_{2}[5x_{1}^{3}]_{0}^{1-x_{2}} dx_{2} = \int_{0}^{\frac{1}{2}} [5x_{2}(1-x_{2})]^{3} dx_{2}\]

Evaluate the integral to find the value of the probability.\[P\left(X_{1}+X_{2} \leq 1\right) = \int_{0}^{\frac{1}{2}} [5x_{2}(1-x_{2})]^{3} dx_{2}\]This would give the final probability that \(X_{1}+X_{2} \leq 1\). Solving this integral would require standard integration techniques.