甲盒内有3个白球与2个黑球，从中任取3个球放入空盒乙中，然后从乙盒内任取2个球放入空盒丙中，最后从丙盒内再任取1个球，试求：(Ⅰ)从丙盒内取出的是白球的概率；(Ⅱ)若从丙盒内取到白球，当初从甲盒内取到3个白球的概率．
【正确答案】正确答案：依题意，有 P(A
1
)=C
3
1
C
2
2
／C
5
3
=3／10，P(A
2
)=C
3
2
C
2
1
／C
5
3
=6／10P(A
3
)=C
3
3
／C
5
3
=1／10； P(B
0
|A
1
)=C
2
2
／C
3
2
=1／3，P(B
0
|A
2
)=P(B
0
|A
3
)=0， P(B
1
|A
1
)=C
2
1
／C
3
2
=2／3，P(B
1
|A
2
)=C
2
2
／C
3
2
=2／3，P(B
1
|A
3
)=0， P(B
2
|A
1
)=0，P(B
2
|A
2
)=1／C
3
2
=1／3，P(B
2
|A
3
)=1． 应用全概率公式 P(B
0
)=
P(A
i
)P(B
0
|A
i
)=P(A
1
)×P(B
0
|A
1
)=1／10， P(B
1
)=P(A
1
)P(B
1
|A
1
)+P(A
2
)P(B
1
|A
2
)
P(B
2
)=1－P(B
0
)－P(B
1
)=3／10， 或P(B
2
)=P(A
2
)P(B
2
|A
2
)+P(A
3
)P(B
2
|A
3
)
P(C|B
0
)=0，P(C|B
1
)=1／2，P(C|B
2
)=1， P(C)=P(B
1
)P(C|B
1
)+P(B
2
)P(C|B
2
)
(Ⅱ)P(C|A
3
)=1，P(A
3
|C)
P(A
i
)P(B
0
|A
i
)=P(A
1
)×P(B
0
|A
1
)=1／10， P(B
1
)=P(A
1
)P(B
1
|A
1
)+P(A
2
)P(B
1
|A
2
)
P(B
2
)=1－P(B
0
)－P(B
1
)=3／10， 或P(B
2
)=P(A
2
)P(B
2
|A
2
)+P(A
3
)P(B
2
|A
3
)
P(C|B
0
)=0，P(C|B
1
)=1／2，P(C|B
2
)=1， P(C)=P(B
1
)P(C|B
1
)+P(B
2
)P(C|B
2
)
(Ⅱ)P(C|A
3
)=1，P(A
3
|C)
【答案解析】