Probability of choosing a defective IC two times

Discussion in 'Homework Help' started by Jess_88, Jun 1, 2011.

  1. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    Hi guys :)
    I'm not sure if this is the right place to post this. I'm assuming there are some people who know about Probability Theory, as some universities include this subject in electrical engineering.

    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?
    [​IMG]

    Where P[A] = probability of selecting box A = 0.5
    P = 0.5

    P[D|X] = probability of selecting two defective IC's
    P[ND|X] = probability of selecting two not defective IC's
    A.D = Two defective IC's from Box A
    A.ND = Two not defective IC's from Box A


    I need some help getting started with this one. I'm really unsure if my tree diagram is correct, mainly due to the selection of two ICs. Is it possible I would use 4 branches from both A and B?

    thanks guys :)
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    If I were you, I would find the probability P1 for two defectives from box A and P2 for two defectives from box B.

    For question 1, I think that it is their average, but I 'm not quite sure.

    As for question two, I don't have a safe answer. Maybe if P1 and P2 add to a whole, their ratio P1/P2 is the answer.
     
    Jess_88 likes this.
  3. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    ok I see.
    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

    dose that look ok?

    I'm also unsure as to my notation. Any corrections would be much appreciated.
     
  4. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    You wrote exactly what I was thinking about problem 1. It seems a bit high, but I don't see a fault in the reasoning and it is certainly lower that box A's alone.

    As for problem 2, I was thinking something like this:
    The chances that the two faulty components are from box A are:
    \frac{1/20}{1/80}=4
    which is 4 to 1
    I like that result, because box A has four times more defective components that box B.
    It is the same result as the one you found.

    I took discrete mathematics 3 years ago, so I can't correct you in any formality error. I simply don't remember them well enough. What I remember though is that it is a very hard field to get accustomed to.

    Good luck with your problem.
     
    Jess_88 likes this.
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,386
    1,605
    Not quite...

    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.
     
    Jess_88 likes this.
  6. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    You are correct. I though about it at some point but then erased it from my mind :p
     
    Jess_88 likes this.
  7. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    I was thinking this as well, but from similar questions given in class, I am assuming both IC's are removed at the same time from the same box.
    So from the tree Diagram, you also need to take into account the odds of choosing box A and box B (50% each). The result of each probability of 2 defective ICs being multiplied by 0.5.

    Now assuming two defective ICs are chosen, and we already know the probability of each. I would think the formula would actually be a little more like Georacer's... what do you guys think?
     
  8. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,386
    1,605
    I don't think that matters, and here is why:

    I pick two IC's from bin A, test the first one and find it is bad. What are the odds the next one is bad? Simply 99/100 as I had said.

    OK, I added some information by testing the first part, so let's see if that is like the Monty Hall problem by doing a simpler case:

    Say I have a bin of 4 parts, and 2 are bad. What is the chance I pick 2 bad ones. This is either:

    2/4 * 2/4 = 1/8
    or
    2/4 * 1/3 = 1/6

    Say I have parts A,B,C and D, where A & B are the bad ones. There are 6 ways to grab 2 parts from 4 parts:

    AB AC AD BC BD CD

    Thus there is only 1 way to pick both bad parts (AB), which is 1 in 6 or 1/6, and I got that without adding any more information to the problem.

    Getting 2 in 8 is the same as putting the first part back before you pick the second one.

    That gives the same result:

    (0.5) * 0.0909 / (0.5) * 0.002261 = 0.0909 / 0.002261 = 4.0202

    Here is doesn't matter, but if I had additional boxes with some defective rate it would matter.

    I play too much winning poker to get things like this wrong: poker is all about applied probability. Bluffing is way overrated, though effective when you give yourself the correct odds to bluff (against a thinking player).

    We have a forum here devoted to just math problems, it might be fun and instructive to post this question there too. I know there are several professional mathematicians who post there.
     
    Jess_88 likes this.
  9. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    Great. Thanks for explaining :).
    yeah I'll post this in the math section as well.
     
  10. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,386
    1,605
    Dang that bertus! Nothin super about HIS moderating!
     
  11. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    HAHAHA. BERTUUUUUS!!! *shaking fist in the air
     
  12. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    A twin thread would be too much, but, Jess, if you want your thread moved in the Math section, send him a PM.

    In general, the Math forum is less populated, but the residents are more involved in the subject.
     
Loading...