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