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m的值的所有可能是1,2,3,4,5,6,n的值的所有可能是1,2,3,4,5,6,
点P(m,n)的所有可能情况有6×6=36种,
(1)点P在圆Q上,即p的坐标满足x2+y2=17,其情况只有P(1,4),P(4,1)两种,
根据古典概型公式,点P在圆Q上的概率为p1=[2/36]=[1/18],
(2)满足x2+y2<17的有(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2)
共有8个,即点P在圆x2+y2=17内部的情况有8种,
由(1)可得,点P在圆Q上只有P(1,4),P(4,1)两种情况,
所以点P在圆Q外部的概率为p2=1-[2+8/36]=[26/36]=[13/18].
点评:
本题考点: 列举法计算基本事件数及事件发生的概率;互斥事件与对立事件.
考点点评: 本题考查等可能事件的概率计算,涉及列举法的运用,注意列举时按一定的顺序,比如,列举点的坐标时,我们把横标从小变大依次列举.
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