SOLUTION: an engineer pulls 30 transistors off the assembly line and 4 are defective. The engineer then randomly selects 8 transistors to test. a. how many different ways can the sample con

Algebra ->  Probability-and-statistics -> SOLUTION: an engineer pulls 30 transistors off the assembly line and 4 are defective. The engineer then randomly selects 8 transistors to test. a. how many different ways can the sample con      Log On


   

Question 929872: an engineer pulls 30 transistors off the assembly line and 4 are defective. The engineer then randomly selects 8 transistors to test.
a. how many different ways can the sample contain AT MOST 3 defective transistors?
B. if at least 2 of the selected transistors are defective, then the engineer will need to stop the production line of why there were more than 2 defective, fix the problem and report. What is the probability that the engineer will need to stop the production?
C. calculate the probability that 3 to 4 defective will be selected
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
30 transistors: 26G, 4def
p(def) = 4/30 = 2/15
...
8 randomly selected
a. how many different ways can the sample contain AT MOST 3 defective transistors?
0 0r 1 0r 2 0r 3 = +%2826C8%29%284C0%29+%2B+%2826C7%29%284C1%29+%2B+%2826C6%29%284C2%29+%2B+%2826C5%29%284C3%29
........
What is the probability that the engineer will need to stop the production?
P(x ≥ 2) = 1 - P(x ≤ 1) = 1 - binomcdf(8, 2/15, 1) = 1 - .71 = .29 0r 29%
P(x= 3 0r x = 4) = binompdf(8, 2/15, 3) + binompdf(8, 2/15, 4) = .065 + .012 = .077
Using a TI calculator 0r similarly a Casio fx-115 ES plus