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Probability question on buckets and balls

  1. Apr 15, 2016 #1
    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

    i am doubting my solution to this problem , therefore i hope someone assists me a bit . it's simple ,
    there is a bucket that contains 5 red balls and 3 white balls , a player picks up 4 balls out of the bucket without returning any . what's the probability of having two red balls and two white balls .
     
    Last edited by a moderator: Apr 15, 2016
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  3. Apr 15, 2016 #2

    Samy_A

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    What is your solution, and how did you derive it?
     
    Last edited: Apr 15, 2016
  4. Apr 15, 2016 #3
    I cannot write this properly on the forum so let A(n,p) stand for an arrangement
    my solution is (A(5,2)*A(3,2))/A(8,4) * 6 , though it gives incorrect results so it's incorrect .
     
    Last edited: Apr 15, 2016
  5. Apr 15, 2016 #4

    Samy_A

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    I can see how you get ##\binom {5}{2} \binom{3}{2}## in the numerator.
    I don't understand the logic behind ##\binom{8}{2}## in the denominator, nor where the 6 comes from (also not clear whether the 6 is supposed to be in the numerator or the denominator).
     
  6. Apr 15, 2016 #5
    6 is in the numerator and sorry that's not A(8,2) i meant to write A(8,4) it's the total number of possible arrangements
    6 -> (r,w,w,r) (w,r,r,w) (r,r,w,w) (w,w,r,r) (r,w,r,w) (w,r,w,r)
     
  7. Apr 15, 2016 #6

    Samy_A

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    Ok, that explains the ##\binom {8}{4}##.
    Well, ##\binom {x}{y}## doesn't assume a specific order, so there is no reason to multiply by 6.
     
  8. Apr 15, 2016 #7
    thank you sir , though i can't just let this one go .. i used two Arrangements not one and then multiplied them together .
    can you please kindly explain to me how can the first format encapsulate all the possibilities ?
    thank you .
     
  9. Apr 15, 2016 #8
    i still can't quite grasp why the 6 is not necessary sir .
    so l'ets say the event wasn't the one i stated and was instead the player get's a red ball a white ball a red ball and then an other white ball , the number of possible combinations for that is A(5,2)*A(3,2) , the thing is that only includes a single order not all of them ,
    can you please explain to me why the 6 is not necessary sir ? the 6 is C(4,2)
     
  10. Apr 15, 2016 #9

    Samy_A

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    As I see it, there are ##\binom{8}{4}## ways to pick 4 balls. That's our denominator.
    How many ways are there to pick 2 red balls? ##\binom{5}{2}##.
    How many ways are there to pick 2 white balls? ##\binom{3}{2}##.
    So the total number of good combinations, those with 2 red balls and 2 white balls, is the product of these two: ##\binom{5}{2}\binom{3}{2}##. That's our numerator.
     
  11. Apr 15, 2016 #10

    Ray Vickson

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    One sure-fire way to do such a problem is to look at all the possibilities; it is not the fastest or most efficient way, but at least it produces correct results.
    First of all, how many (favorable) outcomes are there? These are all the strings of 2Rs and 2Ws, so for example, we can have RRWW or RWRW or RWWR or ... . Here, the string RRWW means the first two are red and the next two are white, etc.

    So, as I said, you can first determine the number of distinct outcomes of that type, recalling that the red balls are all the same as are all the white balls.

    Next, for a typical outcome such as RRWW, what is its probability? What about for RWW? Ditto for RWWR, etc.

    Now just add up all the results.

    BTW: I said this is not the fastest way; but after you do one or two examples this way and notice the pattern, it does, indeed, become the fastest way, or at least as fast as any other method.
     
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