Question 929872
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 = {{{ (26C8)(4C0) + (26C7)(4C1) + (26C6)(4C2) + (26C5)(4C3)}}}
........
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