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Probability difficult bowl of balls

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data

    A bowl contains w white balls. One ball is selected at random from the bowl; its color is noted, and it is returned to the bowl along with n additional balls of the same color. Another single ball is randomly selected from the bowl (now containing w+b-n balls) and it is observed that the ball is black. Show that the (conditional) probability that the first ball is selected was white is [itex] \frac{w}{w+b-n} [/itex]

    (this is exactly how it is written by hand by my prof)

    2. Relevant equations
    I'm using "n" in between A and B as "intersect"

    Well, probability of a, given b, is P(A|B)= [P(AnB)]/[P(B)]
    I think (for independent events) P(AnB)=P(A)xP(B)

    3. The attempt at a solution

    I guess to me it should be w plus b PLUS n, not minus. And I'm not sure how to use conditional probability here, it looks more independent to me. Halp!
     
  2. jcsd
  3. Sep 5, 2011 #2
    Surely you mean that there are w + b + n balls in the bowl the second time, since you added n of them.
     
  4. Sep 5, 2011 #3
    That's what I'm ?? about, but the problem is definitely w+b-n.

    What if there were w+b-2n balls, then when you add n, you have w+b-n. ??
     
  5. Sep 5, 2011 #4
    no, it's a typo from that book you are checking all the problems here.
     
  6. Sep 5, 2011 #5
    Ordinarily I'd agree but this is not from the book, it's hand written problem made by my prof. But he has made mistakes too, so typo is highly possible. I think ill email him.
     
  7. Sep 5, 2011 #6

    I like Serena

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    Hmm, I think I see another typo.
    You write that initially the bowl contains w white balls.
    How could a black ball then be drawn?


    Btw, the equation I think you will need is Bayes' theorem:

    188019d193258f9ba310da979906d24f.png
     
  8. Sep 5, 2011 #7
    Good call on the Bayes theorem, it's one of the key concepts of the homework. As for the balls, I think he's trying to get you to figure out what was in the bowl retroactively.
     
  9. Sep 5, 2011 #8
    Cause it had b black balls.
     
  10. Sep 5, 2011 #9

    I like Serena

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    I give up! :wink:

    This is not a probability problem, it's a find-the-typo problem, or a think-outside-of-the-box problem (since apparently we have to make weird assumptions that the bowl initially contains w white balls accompanied by an unknown number of other balls).
     
  11. Sep 5, 2011 #10
    I know, right? Well I'll update if prof gives a reasonable answer to my email.
     
  12. Sep 5, 2011 #11

    Ray Vickson

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    If W1 = {first ball is white} and B1 = {first ball is black}, what are P{W1} and P{B1}? Now, if {W2} and {B2} refer to the second ball, what are P{B2|W1}, P{B2|B1}, etc? You want P{W1|B2} = P{W1 & B2}/P{B2}. Can you get P{W1 & B2} in terms of P{W1}, P{B1}, P{B2|W1} and P{B2|B1}? Do you know how to get P{B2}?

    By the way: after the first draw the number of balls *is* b + w + n, if your description of the experiment is correct.

    RGV
     
  13. Sep 5, 2011 #12
    Okay, the prof just emailed me back. It is correct as w+b-n. It's due tomorrow. Can I get some more help on it?
     
  14. Sep 5, 2011 #13
    ILS, why do you say Baye's theorem is [tex] P(A|B)= \frac{P(B|A)P(A)}{P(B)} [/tex] ?

    My book says [tex] P(A|B)= \frac{P(B|A)P(A)}{P(A)P(B|A)+P(B)P(B|A)} [/tex]
     
  15. Sep 5, 2011 #14

    Because the denominator of both fractions is the same.
     
  16. Sep 5, 2011 #15
    How's that?
     
  17. Sep 5, 2011 #16
    look at the formula for total probability in your book. It should be in the same section as Bayes's Theorem.
     
  18. Sep 5, 2011 #17
    I tried Baye's theorem and I got w/(w+b)
    This seems kind of reasonable. Since the adding of the n balls occurs AFTER the drawing of the white ball. Now I see why there is -n. But I still have to get that for an answer.
     
  19. Sep 5, 2011 #18

    vela

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    Did you ask your professor to clarify how it could be w+b-n? And for that matter, what does b stand for? Are we supposed to assume it's the number of black balls?
     
  20. Sep 6, 2011 #19
    Arg. After all that, during class today he admits that he wrote the problem inaccurately! Forget it.
     
  21. Sep 6, 2011 #20
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