Question

Suppose that three numbers are selected one by one, at random, and without replacement, from the set of integers {1, 2, 3, . . . , n}. what is the probability that the third number falls between the first two if the first number is smaller than the second?

Answer 1

A - the third number falls between the first two

B - the first number is smaller than the second

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}\\\\ |\Omega|=n(n-1)(n-2)\\\\ |B|=\dfrac{n!}{2}\\\\ |A\cap B|=\dfrac{\dfrac{n!}{2}}{3}=\dfrac{n!}{6}\\\\ P(A\cap B)=\dfrac{\dfrac{n!}{6}}{n(n-1)(n-2)}=\dfrac{n!}{6n(n-1)(n-2)}=\dfrac{(n-3)!}{6}\\ P(B)=\dfrac{\dfrac{n!}{2}}{n(n-1)(n-2)}=\dfrac{n!}{2n(n-1)(n-2)}=\dfrac{(n-3)!}{2}\\\\ P(A|B)=\dfrac{\dfrac{(n-3)!}{6}}{\dfrac{(n-3)!}{2}}\\\\ P(A|B)=\dfrac{(n-3)!}{6}\cdot\dfrac{2}{(n-3)!}=\dfrac{1}{6}=\dfrac{1}{3}$