Please answer fully, clearly, and show work. Need this question within an hour a
ID: 3202019 • Letter: P
Question
Please answer fully, clearly, and show work. Need this question within an hour and a half.
It is known that the defective parts produced in a certain manufacturing process can be made satisfactory by rework or not. It is also known that 30% of these defective parts can be made satisfactory by rework. If a batch of 10 defective parts is selected at random: a. Find the probability that at leas, 3 defective parts can be satisfactorily reworked. b. Find the probability that more than 6 defective parts cannot be satisfactorily reworked. c. Find the probability that exactly 4 defective parts can be satisfactorily reworked, assuming that it is known that 35% of these defective parts can be made satisfactory by reworkExplanation / Answer
P(rework possibility) = 0.3
a) At least 3 can be successfully reworked = 1 - P(0 reworkable) - P(1 reworkable) - P(2 reworkable)
= 1 - (0.7)10 - 10x0.3x0.79 - 10C2x0.32x0.78
= 1 - 0.028 - .0121 - 0.233
= 0.618
b) P (more than 6 cannot be re worked) = 1 - P(7 reworkable) - P(8 reworkable) - P(9 eworkable) - P(10 reworkable)
= 1 - 10C7x0.37x.73 - 10C8x0.38x0.72 - 10x0.39x0.7 - 0.310
= 1 - 0.009 - 0.0014 - 0.00014 - 0.000006
= 0.989
c) P(reworkable) = 0.35
P(exactly 4 out of 10 reworkable) = 10C4x0.354x0.656
= 0.2377