Final answer:
The question covers College-level Mathematics, focusing on probabilities for independent uniform random variables and requires integration over specific regions to find the probabilities P(X>Y) and P(Y^2+X^2>4), while P(Y-X>4) is zero.
Step-by-step explanation:
The student's question involves probability and specifically the calculation of probabilities concerning uniform random variables X and Y in the interval [0,4]. To find P(X>Y), P(Y-X>4), and P(Y^2+X^2>4), it is necessary to understand the properties of uniform distributions, the independence of random variables, and the methods to calculate probabilities for continuous random variables. The situation described implies a two-dimensional uniform distribution where the events are independent, thus the joint probability is the product of the marginal probabilities. However, for P(X>Y), we need to evaluate the integral of the joint probability density function over the region where X is greater than Y. For P(Y-X>4), this probability is zero since the maximum value of Y-X given the specified range [0,4] for both variables would be 4. Lastly, to solve P(Y^2+X^2>4), we again use integration but this time over the region where the sum of the squares of X and Y is greater than 4, which is a circular region in the two-dimensional plane.
When dealing with independent events, such as X and Y, and when calculating probabilities involving inequalities (like X>Y), integrals over the relevant regions of the probability density functions are often used. For the other parts of the question addressed, there are typos that prevent accurate interpretation and calculation.