12/02/18 17:33:53.59
Ｐ{Ａ(m)} = (5/6)^n,
Ｐ{Ａ(i)∩Ａ(j)} = (4/6)^n, (i≠j)
Ｐ{Ａ(i)∩Ａ(j)∩Ａ(k)} = (3/6)^n, (i,j,kは相異なる)
Ｐ{Ａ(1)∩Ａ(2)∩Ａ(3)∩Ａ(4)} = (2/6)^n,
Ｐ{Ａ(1)∩Ａ(2)∩Ａ(3)∩Ａ(4)∩Ａ(5)} = (1/6)^n,
これを確率の加法定理 >>169
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式(17)に入れれば >>166 >>174