a) P(A) = 0.2, P(B) = 0.1, P(none) = 0.7
P(3 no defects, 1 A, and 2 B) =
Hence, probability of 3 no defevts, one type A and two type B defects is 0.08232
b) n = 6, p = 0.2, q = 1 - 0.2 = 0.8
Var(x) = n*p*q = 6*0.2*0.8 = 0.96
I hope these will help. Thank you.
13) Items under inspection are subjected to two types of defects about 70% of the items...
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