2003 CMO 第 5 题

Ten people apply for a job. The manager decides to interview the candidates one by one according to the following conditions:
i) the first three candidates will not be employed;
ii) from the fourth candidates onwards, if a candidate's comptence surpasses the competence of all those who preceded him, then that candidate is employed;
iii) if the first nine candidates are not employed, then the tenth candidate will be employed.
We assume that none of the $10$ applicants have the same competence, and these competences can be ranked from the first to tenth. Let $P_k$ represent the probability that the $k$ th-ranked applicant in competence is employed. Prove that:
i) $P_1>P_2>\ldots>P_8=P_9=P_{10}$ ;
ii) $P_1+P_2+P_3>0.7$ iii) $P_8+P_9+P_{10}\le 0.1$ .

*Su Chun*

十个人申请一份工作。经理根据以下条件决定对应聘者进行一一面试：

i) 前三名候选人将不会被聘用；

ii) 从第四位候选人开始，如果一名候选人的能力超过了所有在他之前的候选人的能力，则该候选人被聘用；

iii) 如果前九名候选人未被聘用，则第十名候选人将被聘用。

我们假设 10 美元的申请者中没有一个具有相同的能力，并且这些能力可以从第一到第十排列。令$P_k$表示能力排名第$k$的申请人被雇用的概率。证明：

i) $P_1>P_2>\ldots>P_8=P_9=P_{10}$ ;

ii) $P_1+P_2+P_3>0.7$ iii) $P_8+P_9+P_{10}\le 0.1$ 。

*苏淳*