22C4264?, 顾客掷得点数小于4,然后抽将得三等奖的概率为?2??3C631515所以P?A??143??; 3155?2?由题意可知,随机变量X的可能取值为100,300,400,
22m?m?1?12Cm1且P?X?100????2??,
33Cm?233?m?2??m?1?11C2C28mP?X?300???2m?,
3Cm?23?m?2??m?1?22C24P?X?400???2?,
3Cm?23?m?2??m?1?所以随机变量X的数学期望,
?12m?m?1??8m4E?X??100????300??400?, ??33?m?2??m?1??3m?2m?13m?2m?1??????????100200m2?2200m?1600?化简得E?X??, 33?m?2??m?1?100200m2?2200m?1600??150, 由题意可知,E?X??150,即33?m?2??m?1?化简得3m2?23m?18?0,因为m?N*,解得m?9, 即m的最小值为9. 【点睛】
本题主要考查概率和期望的求法,属于常考题.
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