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13) Items under inspection are subjected to two types of defects about 70% of the items in a large lot are judged to be defec
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Answer #1

a) P(A) = 0.2, P(B) = 0.1, P(none) = 0.7

P(3 no defects, 1 A, and 2 B) =

\tiny \binom{6}{3}*0.7^3 *\binom{3}{2}*0.1^2 *\binom{1}{1}*0.2 = 0.08232

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.

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