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Probability - Drawing balls from boxes

  1. Jan 30, 2010 #1
    box A has place for 2 balls,
    box B has place for 3 balls

    balls are placed randomly in each of the boxes until one of them is full

    if X is the amount of balls used, find:
    E(x)
    var(x)


    i drew the boxes and think there are only 3 options,,, X=[2 3 4]

    BOX A --- BOX B
    I) 2 balls ---- 0 balls
    II) 2 balls --- 1 ball
    III) 2 balls --- 2 balls
    IV) 1 ball --- 3 balls
    V) 0 balls --- 3 balls

    |__X || 2__|_3__|_4_|
    |P(X) || 1/5 | 2/5 | 2/5|

    E(X)=3.2

    but the correct answer is 3.125

    same problem (obviously) for var
     
  2. jcsd
  3. Jan 30, 2010 #2

    tiny-tim

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    Hi Dell! :smile:

    I think from your "5"s you're misinterpreting "randomly" …

    you're treating it as 5 compartments, each equally likely,

    but it's only 2 boxes, each equally likely. :wink:
     
  4. Jan 30, 2010 #3
    Re: probability.

    not REALLY sure what you think i mean, but the 5 is not from 2+3 (balls) but rather the 5 options i mentioned before, - the 5 RANDOM ways the boxes can be filled
    2-0
    2-1
    2-2
    0-3
    1-3
     
  5. Jan 30, 2010 #4

    tiny-tim

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    ah … but those 5 ways are not equally likely. :wink:

    Try again! :smile:
     
  6. Jan 31, 2010 #5
    Re: probability.

    i thought that might be the problem, but cant fine the probability of each, since the likeliness of choosing either of the 2 boxes is 0.5, i thought the the probabillity of each possibility is 0.5^n (n being the number of balls used)

    2-0 -- 1/4
    2-1 -- 1/8
    2-2 -- 1/16
    0-3 -- 1/8
    1-3 -- 1/16

    but as you can see, this doesnt come to 1. so no point of even continuing
     
  7. Jan 31, 2010 #6

    tiny-tim

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    ok, let's start by doing II) as an example (2A,1B) …

    II) requires the third ball to be in A, and either the first to be in A and the second in B, or vice versa.

    in short, II) is ABA +BAA, so there's two.

    Carry on from there. :smile:
     
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