Question 1198894: Parts are placed in two boxes. The first box contains 8 parts, 6 of them meet
the standard. Second box contains 7 parts, 6 of them meet the standard. From the
first box, three parts are transferred to the second box, and then one part is
taken out from the second box. What is the probability that the part taken out
of the box doesn't meet the standard ?
Answer by Edwin McCravy(20064)   (Show Source):
You can put this solution on YOUR website! Parts are placed in two boxes. The first box contains 8 parts, 6 of them meet
the standard. Second box contains 7 parts, 6 of them meet the standard. From the
first box, three parts are transferred to the second box, and then one part is
taken out from the second box. What is the probability that the part taken out
of the box doesn't meet the standard ?
Case 1: Of the 3 chosen from 1st box, 3 meet the standard, 0 don't.
Probability = (6C3)(2C0)/(8C3) = 20/56 = 5/14
Then 2nd box contains 9 that meet standard, 1 that doesn't
Probability drawing 1 from 2nd box that doesn't meet standard = 1/10
Probability for case 1 = (5/14)(1/10) = 5/140 = 1/28
Case 2: Of the 3 chosen from 1st box, 2 meet the standard, 1 doesn't.
Probability = (6C2)(2C1)/(8C3) = (15)(2)/56 = 15/28
Then 2nd box contains 8 that meet standard, 2 that don't
Probability drawing 1 from 2nd box that doesn't meet standard = 2/10 = 1/5
Probability for case 2 = (15/28)(1/5) = 15/140 = 3/28
Case 3: Of the 3 chosen from 1st box, 1 meets the standard, 2 don't.
Probability = (6C1)(2C2)/(8C3) = (6)(1)/56 = 6/56 = 3/28
Then 2nd box contains 7 that meet standard, 3 that don't
Probability drawing 1 from 2nd box that doesn't meet standard = 3/10
Probability for case 3 = (3/28)(3/10) = 9/280
Final answer = 1/28 + 3/28 + 9/280 = 10/260 + 30/280 + 9/280 = 49/280 = 7/40
Edwin
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