论文部分内容阅读
题 有 6只电器元件 ,其中有 2只次品和 4只正品 ,每次抽取 1只测试后不放回 ,直到两只次品都找到为止 ,求测试 4次能找到两只次品的概率。解 测试的第四次必为次品 ,方法数为C12 ,前 3次必有一次抽到另外一件次品 ,方法数为C13,还有两次抽取 ,抽到的是两件正品 ,方法数为A24
There are 6 electrical components, including 2 defective products and 4 genuine products. Do not put back one test after each test. Until two defective products are found, find the probability of finding two defective products four times. . The fourth time the solution test must be a defective product, the number of methods is C12, and there will be another defective product in the first three times. The number of methods is C13, and there are two extractions. Two genuine products are obtained. The number is A24