Two persons A, B have to speak at a function with 10 other persons. If the persons speak at random order, the probability that A speaks immediately before B is

$\begin{array}{l}\mathrm{Two} \mathrm{persons} \mathrm{A}, \mathrm{B} \mathrm{have} \mathrm{to} \mathrm{speak} \mathrm{at} \mathrm{a} \mathrm{function} \mathrm{with} 10 \mathrm{other} \mathrm{persons}\\ \mathrm{n}\left(\mathrm{S}\right)=12!\\ \mathrm{E}=\text{A speaks immediately before B.}\\ \text{A and B are together in a specific order}\\ \mathrm{n}\left(\mathrm{E}\right)=11!\\ \therefore \mathrm{P}\left(\mathrm{E}\right)=\frac{11!}{12!}=\frac{1}{12}\end{array}$