Probability of one random variable greater than other

Question

Given two random variables x and y with different uniform distribution, i.e., $0\le x \le 1$ and $0.75\le y\le 1$. Find the probability of $x\ge y$ when $x$ and $y$ are chosen randomly.

What I tried was finding probability of $x\le y$ which will be sum of two cases, when $x\le 0.75$ and when $x\ge0.75$: $P(x\le y) = (3/4)(1) + (1/4)(1/2) = 7/8$.

Thus $P(x\ge y) = 1 - 7/8 = 1/8$. Is this right?

Answer 1

If x lands between 0 and 0.75, y will always be greater. $ (3/4)∗(1) $

If x lands between 0.75 and 1, y will be greater 50% of the time, as both are uniform. $ (1/4)*(1/2) $

$ P(x≤y)= (3/4)∗(1)+(1/4)(1/2)=7/8 $

$ P(x≥y)=1−7/8=1/8 $

I agree with your calculations.

Answer 2

It seems right to me, as $P(x\leq y)=P(x\leq 0.75)*P(y\geq 0.75)+P(x\geq 0.75)P(x\leq y|x\geq 0.75)=\frac{3}{4}*1+\frac{1}{4}*\frac{1}{2}=\frac{7}{8}$. $P(x\leq y|x\geq 0.75)=\frac{1}{2}$ due to the fact that $x$, under the assumption that $x\geq 0.75$, and $y$ have the same distribution. Thus you answer is correct.