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Hard conceptual probability problem

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Hello, i encounter a conceptual probability problem that I think i have come into confusio.

    The situation is -

    There are 3 Blue ball and 1 Red ball in a bag (BBBR), while there are 4 Green balls
    in another bag (GGGG) If one of the balls is taken from bag A to bag B (not knowing which colours), then we draw two balls out of bag B (without replacement).......

    what is the probability of drawing two balls of same colour? Another question is - what is the probability of drawing two balls of different colour

    2. Relevant equations



    3. The attempt at a solution


    At first, because I don't know which ball was transferred from bag A to bag B, then I suppose there are four possible situation in bag B. That's GGGGB, GGGGB, GGGGB, GGGGR
    under each possible situation, I set up a formula, (3/4)(2/3), then multiply it by 4.

    But, the possibility is larger than 1, which means it's wrong. But I simply don't understand why.

    How about if, I change the question a little bit. There are one more ball in the first bag, which is Green in colour which becomes BBBRG..

    then take one of the ball from these bags (not specific) to the bag with 4 green balls


    I think it's a hard conceptual problem to solve
     
  2. jcsd
  3. Mar 23, 2012 #2

    NascentOxygen

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    What is exciting about probability questions like this is that you can investigate for yourself to see how close practice is to theory. even before you understand the theory! Get some plastic balls/ conkers/ acorns/ lego blocks/ whatever and two cereal boxes or handy containers and do a hundred trials. Give the boxes a good shake between trials. Record the outcome of every draw so you can process the results. This makes probability exercises come alive. :smile:
     
  4. Mar 23, 2012 #3

    Ray Vickson

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    I have absolutely no idea why you computed (3/4)(2/3)*4, or what that was supposed to represent.

    You have 2 possible bag-B situations: GGGGB (3 choices for B) and GGGGR, so P{GGGGB}=3/4, P{GGGGR} = 1/4. You can even simplify it further: bag B = GGGGN, where N = not-G. Since you are interested only in whether or not the drawn colours are the same it does not matter whether the other colour is B or R; the only thing that matters is that it is not green. However, to convince yourself of this, you might try solving the problem using both of the representations to check if the final answer is the same.

    RGV
     
  5. Mar 23, 2012 #4

    vela

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    Can you explain your thinking behind your attempt? I don't understand what you were doing.
     
  6. Mar 23, 2012 #5





    Thanks two, I am not able to describe why I attempted in that way...but

    RGV, if the first bag, I don't use BBBR, instead I use BBBG, the second bag is still GGGG

    then, what's the probability of getting same balls after two successive (without replacement)

    draw from second bag?
     
  7. Mar 23, 2012 #6

    Ray Vickson

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    You can certainly answer your own question!

    RGV
     
  8. Mar 23, 2012 #7
    but the situation is , the first bag is BBB"G"
     
  9. Mar 23, 2012 #8

    vela

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    So you just randomly wrote some numbers down for no apparent reason and expected it to turn out to be the correct answer? :wink: Surely, you must have had reasons for the the 3/4 and 2/3 and the factor of 4. We need you to explain what you're trying to do or what you're thinking, otherwise we can't see where you're going wrong.
     
  10. Mar 23, 2012 #9

    Ray Vickson

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    Fine. So what is preventing you from listing the possibilities for the second bag after transferring one ball from the first bag? You did it before; why can't you do it now?

    RGV
     
  11. Mar 23, 2012 #10

    HallsofIvy

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    There are three blue and one red ball in the first bag so there is a 3/4 chance that the ball transferred to the second bag is blue (in which case there would be 4 green and one blue ball in the second bag) and a 1/4 chance that the ball transfered to the second bag is red (in which case there would be 4 green and one red ball in the second bag).

    P(X)= P(X|A)P(A)+ P(X|not A)P(not A).
     
  12. Mar 23, 2012 #11
    OK, it's my approach

    If first bag BBBR, second bag GGGG

    then there are four possible situation when one ball transferred into second bag BEFORE
    drawing two balls from second bag

    i.e.

    GGGGB GGGGB GGGGB GGGGR

    For any of the above situation I use the formula

    (3/4) x (2/3) to represent the probability for two green balls successively drawn out.

    then because there are four such situations above but never happens at the same time, so it
    is an "OR" situation

    so I sum up (3/4) x (2/3) + (3/4) x (2/3) + (3/4) x (2/3) + (3/4) x (2/3) which in short
    = (3/4) x (2/3) x 4


    but it's wrong because it's value is larger than 1



    Why I think it should be added up? I am not English user, sorry I don't know how to explain in English effectively, I don't want to mess up things. Or can you tell me why it is wrong to think that the four situation doesn't have to be added up?
     
  13. Mar 23, 2012 #12
    deleted duplicated
     
  14. Mar 23, 2012 #13

    Ray Vickson

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    1) It is not correct to simply sum the probabilities as you have done. Let the event E be E = {draw two green}. We have P{E} = P{E|GGGGB}*P{GGGGB} + P{E|GGGGR}*P{GGGGR}. We have P{GGGGB} = 3/4 and P{GGGGR} = 1/4. Notice that we are adding the probabilities P{E|GGGGB} and P{E|GGGGR}, but must also multiply by the probabilities of those cases occurring; that is, we have (3/4)P{E|GGGGB} + (1/4)P{E|GGGGR}, not P{E|GGGGB} + P{E|GGGGR}.

    2) You computed incorrect probabilities P{E|GGGGB} and P{E|GGGGR}. Can you see why?

    RGV
     
  15. Mar 23, 2012 #14
    Thank you RGV.

    The most difficult thing to me is that I don't know how to determine the situation, not the computation. I am 27 and i've actually learnt these things when I was very young, but now I need it back for reasons of work.

    Then, if you said we can't sum up the probability, but using "conditional probability" instead
    then why , in this case , the probability can be summed up

    let say, three people, A,B,C, the probability of A,B and C solving a problem is 1/2, 1/3, and 1/4 respectively. If i want to find the probability of exactly two of them solving a problem.
    .then why I can add up the probability?

    The solution is - (1/2)(1/3)(1/4) + (1/2)(2/3)(1/4) + (1/2)(1/3)(3/4)
    ( only B and C solve) + ( only A and C solve) + (only A and B solve)

    I can just add the things up. why?

    Oh my god I am fainted
     
  16. Mar 23, 2012 #15

    Ray Vickson

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    Look at the possible contents of bag2 as GGGGB1, GGGGB2, GGGGB3, GGGGR (where we actually identify which blue ball was put into bag 2). Let E = {draw two green from bag 2}. We have E = {E & GGGGB1} union {E & GGGGB2} union {E & GGGGB3} union {E & GGGGR}, so yes, indeed, we add these probabilities:

    P{E} = P{E & GGGGB1} + P{E & GGGGB2} + P{E & GGGGB3} + P{E & GGGGR}.

    Your error was to assume P{E & GGGGB1} = P{E|GGGGB1}, which is NOT TRUE.
    In fact, P{E & GGGGB1} = P{E|GGGGB1}*P{GGGGB1} = (1/4)*P{E|GGGGB1}. All four condtional probabilities are equal, and you do add them up, but you must also divide the sum by 4. Failing to do that was your big mistake. As I said, you also computed P{E|GGGGB1}, etc., incorrectly, and I suggested you go back and do it again. I cannot figure out from your response whether or not you have actually done that.

    The above representation is wasteful: it is easier to just let the contents of bag2 be {GGGGB,GGGGR}, with probabilities P{GGGGB} = 3/4 and P{GGGGR} = 1/4, because we really do not care which blue ball was actually selected. The point, though, is that you can do it either way, provided that you are careful and write things out in detail.

    As I suggested before, you could look instead at bag2 as having contents GGGGN, where N = not green. Now we just have a single term and no conditional probabilities at all: P{E} = prob. of getting two Gs from a bag having 4 Gs and one N.

    RGV
     
    Last edited: Mar 23, 2012
  17. Mar 23, 2012 #16

    NascentOxygen

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    Correct.

    But each of these four bags has equal probability of arising, viz., ¼ or 0.25 or 25%
    That is wrong. For the first pick you are offered 4 greens out of a total of 5 balls.

    So probability of all this: selecting bag #1 and then selecting a pair of greens = ¼ x 4/5 x ¾

    Waiter!! Quickly, glass of water for kenny1999! ZdWS3.gif
     
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