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设Ai=“第i个盖子正好盖在第i个杯子上”,i=1,2,...,5
设B=“至少有一个盖子和杯子配对”
则B=A1+A2+...+A5
P(A1+A2+...+A5)
=P(A1)+P(A2)+...+P(A5)
-P(A1A2)-P(A1A3)-...-P(A4A5)
+P(A1A2A3)-P(A1A2A4)+...+P(A3A4A5)-
...
+P(A1A2A3A4A5)
由于杯子跟盖子配对的方式有5!种,所以
P(A1)=P(A2)=...=P(A5)=4!/5!
P(A1A2)=P(A1A3)=...=P(A4A5)=3!/5!
...
P(A1A2A3A4A5)=1/5!
于是
P(B)
=P(A1+A2+...+A5)
=(4!/5!)*5-(3!/5!)*C(5,2)+(2!/5!)*C(5,3)-...+1/5!
P(全部错误)=1-PB)
=1-(4!/5!)*5+(3!/5!)*C(5,2)-(2!/5!)*C(5,3)+...-1/5!
=(3!/5!)*C(5,2)-(2!/5!)*C(5,3)+...-1/5!
设B=“至少有一个盖子和杯子配对”
则B=A1+A2+...+A5
P(A1+A2+...+A5)
=P(A1)+P(A2)+...+P(A5)
-P(A1A2)-P(A1A3)-...-P(A4A5)
+P(A1A2A3)-P(A1A2A4)+...+P(A3A4A5)-
...
+P(A1A2A3A4A5)
由于杯子跟盖子配对的方式有5!种,所以
P(A1)=P(A2)=...=P(A5)=4!/5!
P(A1A2)=P(A1A3)=...=P(A4A5)=3!/5!
...
P(A1A2A3A4A5)=1/5!
于是
P(B)
=P(A1+A2+...+A5)
=(4!/5!)*5-(3!/5!)*C(5,2)+(2!/5!)*C(5,3)-...+1/5!
P(全部错误)=1-PB)
=1-(4!/5!)*5+(3!/5!)*C(5,2)-(2!/5!)*C(5,3)+...-1/5!
=(3!/5!)*C(5,2)-(2!/5!)*C(5,3)+...-1/5!
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