32212331112P(??1)?C2?(1?)?(1?)?(1?)?(1?)+C2?(1?)?(1?)?(1?)?(1?)4332334421332211, ??(1?)?(1?)?(1?)?(1?)?244332883322122331P(??2)???(1?)?(1?)?(1?)+??(1?)?(1?)?(1?)4433233442321471312 ?C2?(1?)?C2?(1?)?(1?)?44332288P(??3)?1?P(??0，2，4，5)?1?1P(??4)?P(A4)?
3P(??5)?P(A5)?1 8111471197, ?????28828828838288所以：随机变量?的概率分布为：
? P 0 1 11 2882 47 2883 4 1 35 1 28897 2881 811147971110?1??2??3??4??5??. 288288288288383ak?p23.解：（1）由(k?1)ak?1?p(k?p)ak得k?1?p?，k?1，2，3，?，p?1
akk?1故E??0?即
aa24?14?28??4???6，a2??6a1??6；3??4???，a3?16 a12a233a44?3??4???1，a4??16； a34（2）由(k?1)ak?1?p(k?p)ak得即
ak?1k?p?p?，k?1，2，3，?，p?1 akk?1a2ap?1a3p?2p?(k?1)??p?，??p?，?，k??p? a12a23ak?1kak(p?1)(p?2)(p?3)?(p?k?1)?(?p)k?1? a1k!以上各式相乘得∴ak?(?p)k?1?(p?1)(p?2)(p?3)?(p?k?1)
k!(p?1)!(?p)k?1p!k?1?(?p)???
k!(p?k)!pk!(p?k)!1kCp(?p)k，k?1，2，3，?，p 2pk??(?p)k?2?Cp??∴a1?a2?a3???ap ??11112233pp???(1?p)p?1?C(?p)?C(?p)?C(?p)???C(?p)??pppp2?2??? pp