X, Y independent uniform - How to express probabilities

Question

If the arrival time (in minutes) of Alex $X$ is uniform distributed on the interval $[0,75]$ and the arrival time of Jane $Y$ is uniform distributed on the interval $[30, 90]$, $X$ and $Y$ independent find the following probabilities:
a) $P(X<60|X>30)$
b) Calculate the probability only one of the two arrive at the first 60 minutes
c) Calculate the probability at least one of the two arrive at the first 40 minutes
d) Calculate the probability Jane arrives before Alex
How can I express the probability in (b) and (c) ?
$f_X(x)=\dfrac{1}{75},0<x<75$
$f_Y(y)=\dfrac{1}{60},30<y<90$,
$f_{X,Y}(x,y)=f_X(x)f_Y(y)=\dfrac{1}{4500}$, because X and Y independent
For (a) I found $P(X<60|X>30)=\dfrac{P(X<60,X>30)}{P(X>30)}=\dfrac{P(30<X<60)}{P(X>30)}=\dfrac{\int_{30}^{60}f_X(x)dx}{1-P(X<30)}=\dfrac{30/75}{1-\int_0^{30}f_X(x)dx}=\dfrac{30/75}{45/65}=0.6$
For (d) i found $P(Y<X)=1-P(X<Y)=1-(\int_{30}^{90}\int_0^{min(75,y)}f_{X,Y}(x,y)dxdy)=1-0.775=0.225$

Answer 1

I think that some ideas could be the following

b) What you want to evaluate is $$ \mathbb{P}(\{X<60,Y>60\} \cup \{X>60,Y<60\} ) = \mathbb{P}(X < 60, Y > 60) + \mathbb{P}(X>60,Y<60). $$

c) You want to know \begin{align*} \mathbb{P}(\min \{X,Y\} < 40) & = 1- \mathbb{P} (\min \{X,Y\} \ge 40) \\ &= 1 - \mathbb{P} (X\ge 40,Y\ge 40) \\ &= 1 - \mathbb{P} (X\ge 40) \, \mathbb{P} (Y\ge40). \end{align*}