Homework SourseA manufacturer has two factories that make watches A and B D
A manufacturer has two factories that make watches, A and B. Defective watches are made 1 out of 100 times in A, and 1 out of 200 times in B. A retailer receives a large box of watches shipped from one of these factories, but he does not know which one. Suppose he tries a watch from the box and it works. What is the probability that the second watch he tries from the box works, given the first did?
Solution
For Factory A
Probability of defective items P(DA)=1/100
Probability of non-defective itemsP(NDA)=99/100
For Factory B
Probability of defective itemsP(DB)=1/200
Probability of non-defective itemsP(NDB)=199/200
A retailer receives a box from one of the factories A or B
Probability that the box is received from A P(A)=1/2
Probability that the box is received from B P(B)=1/2
Case 1 is; if the retailer receives the box from A
So, Probability that the retailer receives the box from A and it is working(or non defective) is:
P(A and NDA)=P(A) X P(NDA)
=1/2 X 99/100
=99/200 ............................................(C)
Now the next Watch drawn would be working too has the probability 99/100
So P(NDA and C)=P(C) X P(NDA)
=99/200 X 99/100
=0.49005
So the probability that the retailer receives box from factory A and the second one is non defective given that the first is working is 0.49005
Case 2 is; if the retailer receives the box from B
So, Probability that the retailer receives the box from B and it is working(or non defective) is:
P(A and NDB)=P(A) X P(NDB)
=1/2 X 199/200
=199/400 ............................................(D)
Now the next Watch drawn would be working too has the probability 99/100
So P(NDB and D)=P(D) X P(NDB)
=199/400 X 199/200
=0.49501
So the probability that the retailer receives box from factory B and the second one is non defective given that the first is working is 0.49501
The required probability would be
P(either case 1 or case 2)=0.49500+0.49501
=0.98501
THus, the required probability is 98.5%
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