抽到合格产品的事件分别为A、B和C.
(Ⅰ)P(A)=0.90,P(B)=P(C)=0.95.
P(
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| A |
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| B |
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| C |
因为事件A,B,C相互独立,
恰有一件不合格的概率为
P(A•B•
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| A |
=P(A)•P(B)•P(
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| B |
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| A |
=2×0.90×0.95×0.05+0.10×0.95×0.95=0.176
答:恰有一件不合格的概率为0.176;
(Ⅱ)解法一:至少有两件不合格的概率为
P(A•
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| B |
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| C |
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| A |
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| C |
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| A |
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| B |
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| A |
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| B |
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| C |
=0.90×0.052+2×0.10×0.05×0.95+0.10×0.052
=0.012.
答:至少有两件不合格的概率为0.012.
解法二:三件产品都合格的概率为
P(A•B•C)=P(A)•P(B)•P(C)
=0.90×0.952
=0.812.
由(Ⅰ)知,恰有一件不合格的概率为0.176,
所以至少有两件不合格的概率为
1-P(A•B•C)+0.176
=1-(0.812+0.176)
=0.012.
答:至少有两件不合格的概率为0.012.
