Answer:
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.
