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Probabality problem

  1. Feb 15, 2009 #1
    1. box 'a' contains 2 white and 4 black balls, box 'b' contains 5 white and 7 black balls. a ball is transferred from a to b. then a ball is drawn from b. I) what is the probability that it is white? II) and it is black?

    2.find the prob that in a random arrangement of the word UNIVERSITY, 2I's do not come together

    pls help me in solving these problem ASAP.... its urgent
     
  2. jcsd
  3. Feb 15, 2009 #2

    CompuChip

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    Welcome to PF rakesh. Unfortunately you have overwritten that terribly handy template that comes up when you try to post here, so I do not know what you have tried already. But let me give you a hint:
    P(drawing a white ball) = P(drawing a white ball | a white ball was transferred from a to b) + P(drawing a white ball | a black ball was transferred from a to b).

    For the second one, I suggest checking how many rearrangements of the letters there are and how many there are in which the two I's are together.
     
  4. Feb 16, 2009 #3
    Solution:

    box A contains
    2W 4B

    box B contains
    5W 7B

    Step 1)

    When a ball is transferred from A to B

    = For box A: P(w) = Number of chances you might get white (Events)/ Total number chances = 2 / 6 = 1/3

    = For box A: P(b) = 4 / 6 = 2 / 3

    -----------------------------------------------------------------------

    Step 2)

    Now, assuming White is Transferred from A to B
    B contains:
    6W 7B

    So For box B: p(w)= 6 / 13

    Now, assuming Black is Transferred from A to B
    B contains:
    5W 8B

    So For box B: p(b)= 8 / 13
    -----------------------------------------------------------------------

    Step 3:

    Probability of box B is dependent on probability of box A.

    So formula for dependent or probability is

    Total Probability = P(A).P(B|A)

    P(B|A) means Probability of box B Assuming Probability of box A

    So

    Ans
    1) Probablity that is white = Box A P(w) * Box A P(w) = 1/3 * 6/13 = 6/39 = 2/13

    2) Probablity that is black = Box A P(b) * Box A P(b) = 2/3 * 8/13 = 16/39


    is this ok?? pls help
     
  5. Feb 17, 2009 #4

    CompuChip

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    Step 1 and 2 look correct, although your answer is not going entirely as it should :wink:

    Let's look at the probability P(W) that a ball you draw is white. Then
    P(W) = P(W | W) + P(W | B)
    where P(W | A) means: the probability of drawing a white, given that color A (either W(hite) or B(lack)) was transferred. In step 1 and 2 you have reasoned that, for example
    P(W | W) = P(transferring a white ball from A to B) * P(drawing a white ball from B, after a white ball was added) = 2/6 * 6/13
    and
    P(W | B) = P(transferring a black ball from A to B) * P(drawing a white ball from B, after a black ball was added) = 4/6 * 5/13.

    So you get
    P(W) = 2/6 * 6/13 + 4/6 * 5/13 = (6 + 10) / 39 = 16/39
    which seems to be what you have with P(B).

    Also note how your answer cannot be correct: whatever ball you transfer, when you draw from B it must be either black or white. So P(W) + P(B) = 1, which your answer does not satisfy (2/13 + 16/39 = 22/39 =/= 1).

    I think your mistake is in the formula
    P(B) = P(B | A) . P(A)
    which should actually be
    P(B) = Sum[over all possibilities in A]( P(B | A) P(A) )
    so in this case
    P(B) = P(2W,5B in A) P(draw white from B | 2W, 5B in A) + P(3W, 4B in A) P(draw white from B | 3W, 4B in A).
     
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