# Probability that one random variable using the Beta Distribution being greater than another, bounded intervals

I am doing some practice problems to prepare for my statistics exam, and I just want to know if my reasoning is correct on one problem, and if not, I want to know how I should reason through this. The question is as follows:

X, Y are random variables following the Beta distribution B(120, 2019). What is P(2X+4 > 3Y)?


My reasoning is that since X,Y are following the Beta distribution, any x,y $$\in$$ X,Y must be bound to the interval (0,1). Thus, 2X+4 must be bound to (4,6), while 3Y must be bound to (0,3). So would I be right in saying this probability is 100% since min(X)>max(Y)? Thank you for any help. Again, I am not just looking for the answer, I am looking to see if my reasoning is correct, and how I should reason through it if my reasoning is incorrect.

Comment: Your logic seems correct; here is a quick simulation, using R statistical software.

set.seed(2019); m = 10^6
x = rbeta(m, 120, 2019); y = rbeta(m, 120, 2019)
mean(2*x + 4 > 3*y)
[1] 1   # TRUE for 100% of simulated values.


Also, you must have $$D = 2X - 3Y > -4.$$ Here is a summary of the simulated values of $$D,$$ all of which exceed $$-4.$$ Indeed the largest $$D \approx 0.03,$$ which agrees with what you say.

d = 2*x - 3*y
summary(d)
Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
-0.15004 -0.06806 -0.05594 -0.05613 -0.04396  0.03038

• Another, slightly less attractive approach would be to notice that $\mathsf{Beta}(120, 2019)$ is approximately normal, then get a normal approximation of $D$ and show it has (essentially) all its probability above $-4.$ That might be an approach to approximate, say $P(-0.1 < D < -0.05) \approx 0.37,$ without using simulation. – BruceET Jan 21 '19 at 4:15