A - the third number falls between the first two
B - the first number is smaller than the second
[tex] 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} [/tex]
