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Probability color ball problem

  1. Jan 17, 2007 #1
    this question is contained on a gcse past paper i am doing and i think im approaching it from the wrong angle but have tried a few times and dunno where i am going wrong :( hope you can help

    1. The problem statement, all variables and given/known data

    A bag contains (n+7) tennis balls.
    n of the balls are yellow.
    The other 7 balls are white.

    John will take at random a ball from the bag.
    He will look at its colour and then put it back in the bag.

    part a) Bill states that the probability that John will take a white ball is 2/5
    Prove that Bill's statement cannot be correct.

    part b) After John has put the ball back into the bag, Mary will then take at random a ball from the bag.
    She will note its colour.

    Given the probability that John and Mary will take balls with different colours is 4/9, prove that 2n^2 - 35n + 98 = 0

    2. Relevant equations

    3. The attempt at a solution

    a) (I think I got this bit right)

    2/5 = 7/(n+7)
    n+7 = 7/(2/5)
    n+7 = 7/0.4
    n+7 = 17.5
    n = 10.5

    Bill's statement cannot be write because there cannot be a non-interger number of balls.


    I did a probability tree thing and worked out there are 2 ways that they will have different colours, each with a 7n/(n+7) probability.

    therefore, there is a 14n/(n+7) chance of them selecting different colours.

    (this is where i might be going wrong)

    so, 14n/(n+7) = 4/9

    126n/(n+7) = 4
    n+7 = 126n / 4
    n+7 = 31.5n
    7 = 30.5n


    this is where i get stuck cause i realise i shouldnt being working out n should I?

    is it i have to factorise it out somewhere? get it in quadradic form or whatever?

    hope you can help

  2. jcsd
  3. Jan 17, 2007 #2
    I think you had a typo and meant 2n^2 - 3.5n + 98 = 0

    Part (a) looks correct

    Part (b)

    Probability of John or Mary getting yellow : n/n+7
    White : 7/n+7

    Probability of John(yellow) and M(b) = (7/n+7)(n/n+7)
    M(y) J(b) = (n/n+7)(n/n+7)

    Therefore the outcome is going to be the same.

    so expand

    (7/n+7)(n/n+7) = 7n / n^2+14n+49

    Therefore 7n / n^2+14n+49 = 4/9

    63n = 4(n^2 + 14n + 49)
    4n^2 + 56n + 196 -63n = 0
    4n^2 - 7n + 196 = 0 (/2)
    2n^2 - 3.5n + 98 = 0
  4. Jan 17, 2007 #3


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    Science Advisor
    Homework Helper

    Your first step in b) is dubious. 7n/(n+1)>1 for n>1. Can that be a probability? Hint: it maybe clearer to compute the probability they both pick the same color.
  5. Jan 17, 2007 #4


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    Staff Emeritus
    Science Advisor

    Pagey, welcome to PF. Please note that we do not give out full solutions to homework problems, but instead attempt to guide the original poster to the solution.
  6. Jan 17, 2007 #5
    Sorry buddy, never realised, will take note for the future.
  7. Jan 17, 2007 #6
    o rite, yeh i see where i went wrong at begining, stupid error (grrrr)

    i still confused as to how you got ...3.5n... pagey?

    i check and it wasnt a typo
  8. Jan 17, 2007 #7
    i realised where i went wrong, thats utter rubbish.

    The outcome is going to be

    << 2nd complete solution deleted by berkaman >>
    Last edited by a moderator: Jan 17, 2007
  9. Jan 17, 2007 #8


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    Staff: Mentor

    Pagey, you were warned above not to post complete solutions. Stop that! Our task here on the PF is to offer tutorial help and guidance, hints, suggested paths to look down, etc. NOT to do the student's work for them.

    I'm issuing you a warning point. Not a good way to start your tenure at the PF, but hopefully your future posts will be useful in a tutorial way.
  10. Jan 18, 2007 #9
    Sorry dude, but i don't see what i'm supposed to say, i didn't answer, i tried to take note of the fact that i wasn't supposed to answer the question and tryed to provide information that would help him. Sorry for the incovience nevertheless . . .
  11. Jan 18, 2007 #10
    pagey i think you're supposed to say stuff like, "this line should read like this.." and then leave the rest, or say like, "you forgot to do this to that", etc...
  12. Jan 18, 2007 #11


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    Staff Emeritus
    Science Advisor

    If you're unsure as to what constitutes "help," then just take a look at some of the threads in the homework forum and see what help is given to the original poster.
  13. Jan 18, 2007 #12


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    Staff: Mentor

    If you are ending each post with a question, that's a good indicator that you are providing the right kind of help. Having the final answer at the end of your post is a bad indicator.
  14. Jan 18, 2007 #13
    Ok, thanks, i do respect the rules and i think i now understand how to answer properly, thanks for the advice . . .
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