Hi guys 
I posted this question in another section of the forum (homework help) and received some positive feedback. I was also recommended to repost this in the maths section.
So here is my Question.
I have two boxes (A and B), which contain a number of integrated circuits (ICs). Box A has contains 100 ICs, of which 10 percent are defective. Box B has 200 ICs, of which 5 percent are defective.
Two ICs are selected from one randomly selected box.
1) Fined the probability that both ICs are defective.
2) Assuming both ICs are defective, determine the probability that they came from box A.
From my understanding of this question, I would use a tree diagram to help obtain the solutions...correct?
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I was under the impression that I would go about answering the question as follows:
so I have
For box one
P[D|IC1 IC2] = P[D|IC]/N*P[A]
=10/200
=1/20
where P[D|IC1 IC2] = probability of choosing 2 defective IC
P[D|IC] = probability of a defective IC in box A
N*P[A] = number of ICs chosen from the box multiplied by the total number of ICs in the box.
and for box B
P[D|IC1 IC2] = P[D|IC]/N*P
=5/400
=1/80
So the total probability of choosing two defective ICs would be
P[D] = P[AD|IC1 IC2]*P[A|AB] + P[BD|IC1 IC2]*P[B|AB]
where
P[AD|IC1 IC2] = total defective probability of A
P[A|AB] = Probability of choosing box A
P[D] = (1/20)*0.5 + (1/80)*0.5
=1/32
... which seems a little high :s
for question 2.
I am thinking
P[D|A] => 1/32 = (A)1/80 + (B)1/160
1=(A)32/80 + (B)32/160
1-(32/160) = P[D|A]
=0.8
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I was then informed that I should fined the answers as follows (By ErnieM).
1) Box A has 100*.1 = 10 defective IC's The odds of drawing one defective part is just 10/100 = .1 The chance of drawing a second defective part given the first was defective is 9/99=0.0909 (as you already removed 1 defective part, right?).
So the odds of drawing two bad parts is just the product, or 0.00909
Similarly for box B:
(10/200) * (9/199) = 0.002261
2) Just the ratio of probabilities: 0.0909/0.002261= 4.0202 times more likely to get em from A when both are bad.
----------------------------------------------------------------------------
What do you guys think?
also, if anyone is learned in probability theory notation. I would much appreciate any correction in my notation for my attempt at the solution.
I posted this question in another section of the forum (homework help) and received some positive feedback. I was also recommended to repost this in the maths section.
So here is my Question.
I have two boxes (A and B), which contain a number of integrated circuits (ICs). Box A has contains 100 ICs, of which 10 percent are defective. Box B has 200 ICs, of which 5 percent are defective.
Two ICs are selected from one randomly selected box.
1) Fined the probability that both ICs are defective.
2) Assuming both ICs are defective, determine the probability that they came from box A.
From my understanding of this question, I would use a tree diagram to help obtain the solutions...correct?
----------------------------------------------------------------------------
I was under the impression that I would go about answering the question as follows:
so I have
For box one
P[D|IC1 IC2] = P[D|IC]/N*P[A]
=10/200
=1/20
where P[D|IC1 IC2] = probability of choosing 2 defective IC
P[D|IC] = probability of a defective IC in box A
N*P[A] = number of ICs chosen from the box multiplied by the total number of ICs in the box.
and for box B
P[D|IC1 IC2] = P[D|IC]/N*P
=5/400
=1/80
So the total probability of choosing two defective ICs would be
P[D] = P[AD|IC1 IC2]*P[A|AB] + P[BD|IC1 IC2]*P[B|AB]
where
P[AD|IC1 IC2] = total defective probability of A
P[A|AB] = Probability of choosing box A
P[D] = (1/20)*0.5 + (1/80)*0.5
=1/32
... which seems a little high :s
for question 2.
I am thinking
P[D|A] => 1/32 = (A)1/80 + (B)1/160
1=(A)32/80 + (B)32/160
1-(32/160) = P[D|A]
=0.8
-----------------------------------------------------------------------------------
I was then informed that I should fined the answers as follows (By ErnieM).
1) Box A has 100*.1 = 10 defective IC's The odds of drawing one defective part is just 10/100 = .1 The chance of drawing a second defective part given the first was defective is 9/99=0.0909 (as you already removed 1 defective part, right?).
So the odds of drawing two bad parts is just the product, or 0.00909
Similarly for box B:
(10/200) * (9/199) = 0.002261
2) Just the ratio of probabilities: 0.0909/0.002261= 4.0202 times more likely to get em from A when both are bad.
----------------------------------------------------------------------------
What do you guys think?
also, if anyone is learned in probability theory notation. I would much appreciate any correction in my notation for my attempt at the solution
.
