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5 August, 16:11

In each lot of 100 items, two items are tested, and the lot is rejected if either of the tested items is found defective. (a) Find the probability that a lot with k defective items is accepted. (b) Suppose that when the production process malfunctions, 50 out of 100 items are defective. In order to identify when the process is malfunctioning, how many items should be tested so that the probability that one or more items are found defective is at least 99%?

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  1. 5 August, 19:30
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    a) The probability that a lot is accepted is 1 + (k²-199k) / 9999

    b) It should be tested with at least 7 items to obtain that 99% of the time at least one item is tested.

    Step-by-step explanation:

    a) The probability that the first item is non defective is

    100-k/100

    From the 99 items remaining, the probability that the second item is also non defective is

    99-k/99

    Thus, the probability that the lot is accepted is

    100-k/100 * 99-k / 99 = (9900 - 199k + k²) / 9900 = 1 + (k²-199k) / 9999

    b) If 50 out of 100 are defecitve, then if we test only 2, the probability that the lot is accepted is

    1 - (199*50+2500) / 9999 = 0.2549

    Pretty high. If we prove with a third item, the probability thay the three items pass the test is

    0.2549 * (98-50) / 98 = 0.1248

    If we prove with 4, it is

    0.1248 * 47/97 = 0.0605

    If we prove with 5, it is

    0.0605*46/96 = 0.03

    If we prove with 6:

    0.03*45/95 = 0.013

    And if we prove with 7, it is

    0.013 * 44/94 = 0.00642

    Since this number is less than 0.01, then if we test 7 items, we will find that in more than 99% of the cases at least one is defective.
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