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Bayes theorem for Probability

  1. Jan 23, 2010 #1
    1. At an electronics plant, there is an optional training program for new
    employees. From past experience it is known that 87% of new employees
    who attend the training program meet the production quota in the first week
    of work. It is also known that only 34% of workers who do not attend the
    training program meet the production quota in the first week of work. They
    also know that 82% on new employees attend the training program.
    (a) What percentage of new employees will meet the production quota in
    their first week of work?
    (b) If a new employee did meet the production quota in their first week of
    work, what is the probability that they did not attend the training
    program?
    (c) If a new worker is selected at random, what is the probability that they
    did not meet the production quota and/or they did not attend the
    training program?.








    3. Ok basically, from the info given P(B|A)=.87, P(B|A^c)=.34 and P(A)=.82 where A=(attends the training) and B=(meets production quota in first week) where ^c are the compliments.

    a) i got P(B) from P(B)=P(B n A)+P(B n A^c) which i put into a dependent product P(B|A)P(A)+... = .9922 (this value seems to high to me? what am i doing wrong)

    also in part b) it asks for P(A|B) so i used bayes theorem since i knew P(B|A) and got .719?

    c) i have no idea how to do can someone tell me what they are even asking! Please help i am so confused!!!
     
  2. jcsd
  3. Jan 23, 2010 #2

    sylas

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    Re: Probability

    Method for (a) is good, but the answer is wrong. Show a bit more working and we'll see.

    Method for (b) is good, but the answer is wrong. What is Bayes theorem?

    For (c) they are asking for the probability that they failed quota or they missed training. The ideal employee attends the training and then meets quota. SO they are asking how many employees are not in this ideal.

    Cheers -- sylas
     
  4. Jan 23, 2010 #3
    Re: Probability


    why is it wrong? i dont understand.. you dont know what bayes thm is? can u show me how u got the answer and what did u get and do different from me
     
  5. Jan 23, 2010 #4

    vela

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    Re: Probability

    Because it's not the right numeric answer. Unfortunately, you didn't show enough work to indicate where you went awry.

    It doesn't work that way. You're supposed to show what you did in sufficient detail so others can help you identify where the problem is.
     
  6. Jan 23, 2010 #5

    sylas

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    Re: Probability

    Sorry, my question was unclear. I am not asking because I don't know; but because I want to help you work through the problem step by step. As vela points out, you need to show more about how you got your answer before we can give out my answers. That's just the way we manage questions like this one.

    Bayes theorem is a useful way to calculate Pr(A|B) given Pr(B|A), as you say. So if you can write down all the steps you are applying, including your own statement of the theorem, then we can help you fix up any problems.

    Cheers -- sylas
     
  7. Jan 23, 2010 #6
    Re: Probability


    wouldnt be alot easier if we could compare answers? i'm not doing this to copy answers that is stupid its for understanding and right now i don't know if i'm right or wrong how can i and where do i go?

    first off for a) i want P(B)

    thus i used property: P(B) = P(B n A)+ P(B n A^c)
    then i broke it down into P(B|A)P(A) + P(B|A^c)P(A) since A/B are dependent.
    then i just plugged in value of P(B|A) its complement and P(A)
    and got .9922 i used all the properties correctly... where did i go wrong?

    if i got part a) wrong i get part b) wrong because it carries over but im not sure where i went wrong?
     
  8. Jan 24, 2010 #7

    vela

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    Re: Probability

    Well, you already said you think your answer is wrong, and we confirmed that. I'm not sure why you are so resistant to showing your actual calculation. It's literally just one line.

    So you said P(B)=0.87*0.82+0.34*0.82? That's a bit strange because it doesn't work out to 0.9922, which is what you said you got. It's equal to 0.9758 (which is also wrong).

    You apparently didn't, and again, it's impossible to identify what you did wrong because you refuse to tell us specifically what you did.
     
  9. Jan 24, 2010 #8

    sylas

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    Re: Probability

    This is correct: "P(B) = P(B n A)+ P(B n A^c)"

    This is where you went wrong: "then i broke it down into P(B|A)P(A) + P(B|A^c)P(A)"

    The first half used "P(B n A) = P(B|A)P(A)" which is good.
    The second half used "P(B n A^C) = P(B|A^c)P(A)" which is wrong. The P(A) in this needs to be something else.

    When you fix that, you will have got (a) correct.

    Stick with it. This is how we do it. You show your own working, and we help you fix up the working. In the end, you'll be the one presenting the right answer. You're almost at that point now.

    Cheers -- sylas
     
  10. Jan 24, 2010 #9
    Re: Probability


    oops it should be P(B|A^C)P(A^C)
    thanks.
    also for these types of problems how do you KNOW the difference between P(A n B) and P(A|B) or P(B|A) they sound similar. any hints for differentiating between the two? this case it sounded more like a conditional one because "given" they took the optional training they met the quota in the first week is what it sounded like it could have been used with an "and" statement too?
     
    Last edited: Jan 24, 2010
  11. Jan 24, 2010 #10

    sylas

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    Re: Probability

    Bingo. That should fix it.

    I think it depends on the person what helps them remember the differences better. The words are enough for me, and I suspect this comes with a bit of practice.

    I think of it like this, I guess. "A and B" is a stronger condition than just "A". So it has to have a probability less than or equal to just "A" by itself, or just "B" by itself.

    "A given B" is a different set of circumstances, which might make A more likely or less likely. I'm not being asked for a probability of B occurring at the same time as A. I'm TOLD that B has occurred already, and given this, how likely is A? It might be more, or less, than P(A) given no extra information.

    But what helps you remember might be different.

    Cheers -- sylas
     
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