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%?

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.

29 is the correct answer

step-by-step explanation:

idk

step-by-step explanation: