若把连续掷两次骰子分别得到的点数m、n作为点P的坐标,则点P落在圆x2+y2=25外的概率是(  )

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  • 连续抛掷两次骰子分别得到的点数m,n作为点P的坐标所得P点有:

    (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

    (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

    (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

    (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

    (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

    (6,1),(6,2),(6,3),(6,4),(6,5),(6,6).共36个

    其中落在圆x2+y2=25外的点有:

    (1,5),(1,6),(2,5),(2,6),

    (3,5),(3,6),(4,4),(4,5),(4,6),

    (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

    (6,1),(6,2),(6,3),(6,4),(6,5),(6,6).共21个

    故点P落在圆x2+y2=25外的概率P=[21/36=

    7

    12]

    故答案为 [7/12]