Uniform Random Number

Question

Two uniform random numbers are chosen one after the other. what is the probability of second number second random number greater than first number? I tried this way Please correct me if I am wrong.

$$X_1 \sim U(20,30)\\ X_2 \sim U(20,30)\\ P(X_2\geq X_1) = P(X_2-X_1\geq0) = 1-P(X_2-X_1 \geq 0) = 1- \left[X_3 - \frac{20}{10}\right].$$

I am stuck after this. I also tried solving this using conditional probability

$$P(X_2 > X_1 | X1) = \left(30-\frac{X_1}{10}\right)$$

Please help.

Thanks, Ram

Answer 1

By symmetry, there can be no other probability other than $\frac12$. The probability of the two numbers being equal is $0$, and the probability of the first bein bigger is the same as the probability of the first being smaller.

Answer 2

Let $X$ and $Y$ be iid random variables. Then $$P(X<Y)=P(Y<X)$$ Next to that we have: $$1=P(X<Y)+P(X=Y)+P(Y<X)$$ This together leads to: $$P(X<Y)=P(Y<X)=\frac{1}{2}(1-P(X=Y))$$ In your case $P(X=Y)=0$ so we end up with $$P(X<Y)=P(Y<X)=\frac{1}{2}$$