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北京市西城区2007年5月份抽样测试题的第15题,曾先后在多种出版物上出现,其不同的版本上的解法各不相同,为避免该题解答的混乱状况,现就此题以及此类问题的不同解法进行分析.问题有6件产品,其中含有3件次品,现逐个抽取检查(不放回),求:(Ⅰ)前4次恰好查出两件次品的概率;(Ⅱ)设查出全部次品时检查产品的个数为ζ,求ζ的分布列与数学期望.解答:(Ⅰ)P=(C_3~2C_3~2A_4~4)/A_6~4=3/5.第(Ⅰ)问的解法没有问题.以下就第(Ⅱ)问的不同解法进行分析.解法1:有两种出版物上的解法如下:当ζ=3时,即在6次抽查中,前3次就查出全部3件次品,或前3次查出全部3件正品,均视为检查出全部3件次品,∵P(ζ=3)=A_3~3/A_6~3×2=1/10;同理,当ζ=4、5时,有P(ζ=4)=(C_3~2A_3~3C_3~1)/A_6~4×2=3/10;P(ζ=5)=(C_4~2A_3~3C_3~2A_2~2)/A_6~5×2=3/5;∴ζ的分布列为
The 15th question of the May 2007 sampling test in Xicheng District of Beijing Municipality has appeared in various publications successively. The solutions to the different versions are different. To avoid confusion in the resolution of the question, the question is now The different solutions to this type of problem are analyzed. The problem is that there are 6 products, which contain 3 defective products. Now we are going to take the inspections one by one (without putting back), and find: (I) the probability of finding two defective products exactly 4 times before; (II) Set the number of products to be checked when all defective products are detected to ζ, and find out the distribution of the defects and the mathematical expectation. Answer: (I) P=(C_3~2C_3~2A_4~4)/A_6~4=3/ 5. There is no problem with the solution to question (I). The following analysis of the different solutions to question (II) is given. Solution 1: There are two solutions on the following publications: When ζ = 3, that is, in the six random inspections. In the first three times, all three defective products were found, or all three genuine products were found in the first three times, and all three defective products were deemed to be inspected. ∵P(ζ=3)=A_3~3/A_6~3 ×2=1/10; Similarly, when ζ=4, 5, P(ζ=4)=(C_3~2A_3~3C_3~1)/A_6~4×2=3/10; P(ζ= 5) = (C_4~2A_3~3C_3~2A_2~2)/A_6~5×2=3/5; the distribution of ∴ζ is listed as