Probably $x_1 \leq x_3 \leq x_2$ From set  ${1,2...n}$ we choose three random different numbers, $x_1, x_2,x_3$. If $x_1 \le x_2$, find probability $x_1 \le x_3 \le x_2$.
I set A is event  $x_1 \leq x_3 \leq x_2$, and B $x_1 \leq x_2$, and find P(AB)=$\frac{1}{6}$, P(B)=$\frac{1}{2}$. Is it okay?
 A: The problem has two different solutions, depending on whether you take them with replacement or without it (your question is quite ambiguous in that sense: the word 'different' suggests that they are taken without replacement, and the sign '$\leq$' - that they are taken with replacement):
Solution for numbers taken without replacement:
$$P(x_1 < x_2) = \frac{1}{2}$$
$$P(x_1 < x_3 < x_2) = \frac{1}{6}$$
$$P(x_1 < x_3 < x_2| x_1 < x_2) = \frac{1}{3}$$
Solution for numbers taken with replacement:
$$P(x_1 \leq x_2) = P(x_1 = x_2) + P(x_1 < x_2) = P(x_1 = x_2) + \frac{1 - P(x_1 = x_2)}{2} = \frac{1}{n} + \frac{n-1}{2n} = \frac{n+1}{2n}$$
$$P(x_1 \leq x_3 \leq x_2) = P(x_1 < x_3 < x_2) + P(x_1 = x_3 < x_2) + P(x_1 < x_3 = x_2) - P(x_1 = x_3 = x_2) = \frac{1 - P(x_1 = x_3 < x_2) - P(x_1 < x_3 = x_2) + P(x_1 = x_3 = x_2) }{6} + P(x_1 = x_3 < x_2) + P(x_1 < x_3 = x_2) - P(x_1 = x_3 = x_2) = \frac{1}{6} + \frac{5}{6}P(x_1 = x_3 < x_2) + \frac{5}{6}P(x_1 < x_3 = x_2) - \frac{5}{6}P(x_1 = x_3 = x_2) = \frac{1}{6} + \frac{5(n+1)}{12n^2} + \frac{5(n+1)}{12n^2} - \frac{5}{6n^2} = \frac{n+5}{6n}$$
$$P(x_1 \leq x_3 \leq x_2| x_1 \leq x_2) = \frac{n+5}{6n}\frac{2n}{n+1} = \frac{n+5}{3(n+1)}$$
