1) In a batch of 48 shafts, there are 8 shafts that do not meet the length speci
ID: 3049161 • Letter: 1
Question
1) In a batch of 48 shafts, there are 8 shafts that do not meet the length specifications. A sample of 4 shafts are selected at random (without replacement) and inspected.
What is the probability that there are exactly 2 shafts in the sample that does not meet the length specifications?
2) In a batch of 41 shafts, there are 7 shafts that do not meet the length specifications. A sample of 4 shafts are selected at random (without replacement) and inspected.
What is the probability that there are at least two shafts in the sample that do not meet the length specifications?
3) In a batch of 89 shafts, there are 6 shafts that do not meet the length specifications. A sample of 11 shafts are selected at random (without replacement) and inspected.
What is the mean number of shafts in the sample that do not meet the length specifications?
Explanation / Answer
#1.
4 shafts from 48 shafts can be selected in 48C4 ways.
2 shafts with defect from 8 can be selected in 8C2 ways and remaining 2 non defective can be selected in 40C2
Hence required probability = (8C2 * 40C2)/48C4 = 0.1122
#2.
4 shafts from 41 shafts can be selected in 41C4 ways.
Less 2 shafts with defect from 7 can be selected in (7C0 + 7C1) ways and remaining non defective can be selected in (34C4 + 34C3)
probability of selecting less than 2 defective shaft is (7C0*34C4 + 7C1*34C3)/41C4 = 0.8716
Hence required probability = 1 - 0.8716 = 0.1284
#3.
ratio of defective shafts i.e. p = 6/89 = 0.0674
Hence mean number of shafts that do not meet the specifications = 11*0.0674 = 0.7416
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