1221C2gC2?C2gC13且在一局游戏中得2分的概率为; ?3C51021C2gC21则P(X?1)??, 3C55P(X?2)?P(X?3)?P(X?4)?436, ??5102543228, ?(1?)??510512543342, ?(1?)??5105125?X的分布列为:
X P E(X)?1?1 1 52 6 253 28 1254 42 125162842337. ?2??3??4??525125125125
1221C2gC2?C2gC13且在一局游戏中得2分的概率为; ?3C51021C2gC21则P(X?1)??, 3C55P(X?2)?P(X?3)?P(X?4)?436, ??5102543228, ?(1?)??510512543342, ?(1?)??5105125?X的分布列为:
X P E(X)?1?1 1 52 6 253 28 1254 42 125162842337. ?2??3??4??525125125125