4. At a factory, three robots (A, B and C) build devices, 20% of the devices mad

Question

4. At a factory, three robots (A, B and C) build devices, 20% of the devices made by A are defective. 10% of the devices made by B are defective, and 5% of the devices made by machine C are defective. Each robot builds the same number of devices per day. Let F indicate a defective device. (a) What s the probability P(F) that a given device produced in this factory is defective? (b) Given that a device is defective, what is the probability that it was produced by robot A? (c) Given that a device is not defective, what is the probability that it was produced by robot C?

Explanation / Answer

a) P(F) = P(F | A)*P(A) + P(F | B)*P(B) + P(F | C)*P(C)

P(F) = 0.20*(1/3) + 0.10*(1/3) + 0.05*(1/3)

P(F) = 7/60

P(F) = 0.12 [Rounded off to 2 decimal places]

b) P(A | F) = P(A and F)/ P(F)

P(A | F) = (1/3* 0.20)/0.12

P(A | F) = 4/7

c) P(Not defective) = 1 - P(Defective) = 1 - 7/60 = 53/60 = 0.88

Hence,

P(C | Fc) = P(C and FC) / P(FC)

P(C | Fc) = (1/3 * 0.05)/0.88

P(C | Fc) = 1/60

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