Each day, I receive a shipment of three parts from one of three warehouses, wher

Question

Each day, I receive a shipment of three parts from one of three warehouses, where each of the three warehouses is equally likely to have been the source of that shipment. • Given the shipment came from Warehouse 1, the status of each part is conditionally independent of the status of other parts in that shipment, and there is a probability of 0.5 that a given part is defective. • Given the shipment came from Warehouse 2, the status of each part is conditionally independent of the status of other parts in that shipment, and there is a probability of 0.3 that a given part is defective. • Given the shipment came from Warehouse 3, the status of each part is conditionally independent of the other parts in that shipment, and there is a probability of 0.1 that a given part is defective. (a) Suppose that I receive a shipment. What is the probability that the first part in that shipment is defective? (b) Suppose that I receive a shipment. What is the probability that all three parts in that shipment are defective? (c) Suppose that I receive a shipment and note that all three parts are defective. What is the probability that the shipment came from Warehouse 1? (d) Suppose that I receive a shipment and the first two parts are defective. What is the probability that the third part is also defective?

Explanation / Answer

let shipment from warehouse probability =P(W1)=1/3, from warehouse 2 P(W2)=1/3 and from warehouse 3 probabilty P(W3)=1/3

also produict is defective given from warehouse 1 =P(D|W1) =0.5

produict is defective given from warehouse 2 =P(D|W2) =0.3

produict is defective given from warehouse 1 =P(D|W3) =0.1

hence probabilty of defective =P(D)=P(W1)(D|W!)+P(W2)(D|W2)+P(W3)(D|W3) =(1/3)(0.5+0.3+0.1) =0.3

a)probability that the first part in that shipment is defective =P(D) =0.3

b) probability that all three parts in warehouse 1 are defective =P(3D|W1) =0.53 =0.125

probability that all three parts in warehouse 2 are defective=P(3D|W2) =0.33 =0.027

probability that all three parts in warehouse 3 are defective=P(3D|W3) =0.13 =0.001

hence probabilty of 3 defective =P(3D)=P(W1)(3D|W!)+P(W2)(3D|W2)+P(W3)(3D|W3) =(1/3)(0.125+0.027+0.001)

=0.051

c)  probability that the shipment came from Warehouse 1, given all 3 parts are defective =P(W1|3D)

=P(W1)(3D|W!)/P(3D) =(1/3)*(0.125)/0.051=0.817

d) probabilty of 2 defective =P(2D) =(1/3)(0.52+0.32+0.12) =0.1167

hence  probability that the third part is also defective given first 2 are defective =P(D3|D2) =P(D3)/P(D2)

=0.051/0.1167=0.437

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