Four machines: A, B, C and D produce respectively: 60%, 15%, 10% and 15% of the

Question

Four machines: A, B, C and D produce respectively: 60%, 15%, 10% and 15% of the total number of articles of a factory. The percentages of production defects of these machines are: 0.5%, 0.6%, 0.7% and 0.4% respectively; suppose that an item is selected at random and turns out to be non-defective, what is the probability that the item was produced by machine B? Four machines: A, B, C and D produce respectively: 60%, 15%, 10% and 15% of the total number of articles of a factory. The percentages of production defects of these machines are: 0.5%, 0.6%, 0.7% and 0.4% respectively; suppose that an item is selected at random and turns out to be non-defective, what is the probability that the item was produced by machine B?

Explanation / Answer

X - defective

P(B|X')

= P(B and X') / P(X')

P(X') = P(X'|A)P(A) + P(X'|B)P(B) + P(X'|C)P(C) + P(X'|D)P(D)

= 0.60 * ( 1 - 0.005) + 0.15 * (1 - 0.0006) + 0.10 * (1 - 0.007) + 0.15 *(1 - 0.004)

= 0.99561

P(B and X') = P(X'|B)P(B) = 0.15 * (1 - 0.0006) = 0.14991

hence required probability = 0.14991/0.99561

= 0.150571

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