Answer:
[tex]P(Y>X) = \frac{17}{32}[/tex]
Step-by-step explanation:
Recall that since X is uniformly distributed over the set [1,4] we have that the pdf of X is given by [tex]f(x) = \frac{1}{3}[/tex] if [tex]1\leq x \leq 4[/tex] and 0 otherwise. In the same manner, the pdf of Y is given by [tex]g(y) = \frac{1}{4}[/tex] if [tex]1\leq y\leq 5[/tex] and 0 otherwise.
Note that if Y is in the interval (4,5] then Y>X by default. So, in this case we have that P(Y>X| y in (4,5]) = 1. We want to calculate the probability of having Y in that interval . That is
[tex]P(Y>4) = \int_{4}^{5}\frac{1}{4}dy = \frac{1}{4}[/tex]. Thus, [tex]P(Y\leq 4 ) = 1 - P(Y>4)= \frac{3}{4}[/tex].
We want to proceed as follows. Using the total probability theorem, given two events A, B we have that
[tex]P(A) = P(A|B)\cdot P(B) + P(A|B^c) \cdot P(B^c)[/tex] In this case, A is the event that Y>X and B is the event that Y is in the interval (4,5].
If we assume that X and Y are independent, then we have that the joint pdf of X,Y is given by [tex]h(x,y) = f(x)g(y) = \frac{1}{12}[/tex] when [tex]1\leq x \leq 4, 1\leq y \leq 4[/tex]. We can draw the region were Y>X and the function h(x,y) is different from 0. (The drawing is attached). This region is described as follows: [tex]1\leq x \leq 4[/tex] and [tex]x\leq y \leq 4[/tex], then (the specifics of the calculations of the integrals are ommitted)
[tex]P(Y>X| y \in [1,4)) = \int_{1}^{4}\int_{x}^{4}\frac{1}{12}dy dx = \frac{1}{12}\int_{1}^{4} (4-x) dx = \frac{1}{12}\left.(4x-\frac{x}{2})\right|_{1}^{4}= \frac{1}{12}(4\cdot 4 - \frac{4^2}{2}-(4-\frac{1}{2}) = \frac{9}{2\cdot 12} = \frac{3}{8}[/tex]
Thus,
[tex]P(Y>X) = 1\cdot \frac{1}{4} + \frac{3}{8}\cdot \frac{3}{4} = \frac{17}{32}[/tex]
